Q: Calculate (a) molality (b) molarity and(c) mole fraction of \$KI\$ if the density of \$20\%\$ (mass/ mass) aqueous \$KI\$ is \$1.202\ g\ mL^{-1}\$

Molality \$ \dfrac{w}{w}\% = 20 \$ \$100\ gm \$ solution has \$20\ gm \$ KI \$80\ gm \$ solvent has \$20\ gm \$ KI \$ m = \dfrac{\dfrac{20}{166}}{\dfrac{80}{1000}} = 1.51 mol \$ Molarity Volume = \$\dfrac{m}{d} \Rightarrow \dfrac{20}{1.2} = 83.2 \$ Molalrity \$ = \dfrac{.120 \times 1000}{83.2} \$ \$ = 1.44\ m \$ Mole fraction = \dfrac{\dfrac{20}{166}}{\dfrac{80}{18} + \dfrac{20}{166}}

Q: Calculate the morality of \$NaOH\$ in the solution prepared by dissolving its \$4\ g\$ in water to make solution of \$250\ ml\$.

Weight of \$NaOH =4 g\$ Molecular weight of \$NaOH =40 G\$ Volume of solution is \$250ml\$ Molarity = ? \$M=\frac{W}{MW}\times \frac{1000}{V in ml}\$ \$M=\frac{4}{40}\times \frac{1000}{250}\$\$0.4M\$

Q: 50 kg of N2(g) and 10kg of \$H_2(g)\$ are mixed to produce \$NH_3 (g)\$ . Calculate the amount of \$NH_3\$ formed. Identify the limiting reagent in the production of the \$NH_3\$.

As we know that,\$\text{no. of moles} = \cfrac{\text{Wt.}}{\text{mol. wt.}}\$Weight of \${N}_{2} = 50 \; kg\$Molecular weight of \${N}_{2} = 28 \; g\$No. of moles of \${N}_{2} = \cfrac{50 \times {10}^{3}}{28} = 17.86 \times {10}^{2} \text{ moles}\$Weight of \${H}_{2} = 10 \; kg\$Molecular weight of \${N}_{2} = 2 \; g\$No. of moles of \${N}_{2} = \cfrac{10 \times {10}^{3}}{2} = 5 \times {10}^{3} \text{ moles}\$\${N}_{2} + 3 {H}_{2} \longrightarrow 2 N{H}_{3}\$From the above reaction,\$1\$ mole of \${N}_{2}\$ react with \$3\$ moles of \${H}_{2}\$.No. of moles of \${H}_{2}\$ required to react with \$17.86 \times {10}^{2}\$ moles of \${N}_{2} = 3 \times 17.86 \times {10}^{2} = 5.36 \times {10}^{3} \text{ moles}\$But only \$5 \times {10}^{3}\$ moles of \${H}_{2}\$ are available.Thus \${H}_{2}\$ is the limiting reagent here.Now, again from the above reaction,Amount of ammonia formed when \$3\$ moles of \${H}_{2}\$ react \$= 2 \text{ moles}\$Therefore,Amount of ammonia formed when \$5 \times {10}^{3}\$ moles of \${H}_{2}\$ react \$= \cfrac{2}{3} \times \left( 5 \times {10}^{3} \right) = 3.33 \times {10}^{3} \text{ moles}\$Molecular weight of ammonia \$= 17 \; g\$Weight of ammonia in \$3.33 \times {10}^{3}\$ moles \$= 17 \times \left( 3.33 \times {10}^{3} \right) = 56.61 \; kg\$.Hence the amount of \$N{H}_{3}\$ formed is \$56.61 \; kg\$.

Q: The molarity of pure water is:

Formula:\$Molarity=\cfrac { No\ of\ moles\ of\ solute }{ Volume\ of\ solution\ in\ litre } \$Step 1: To find the mass of water in one liter.We know that,Density of water = \$1 g/ml\$ or = \$ 1 kg/L\$Therefore, the mass of \$ 1 L \$ of water will be:\$ Mass =\ Density\ of\ water\ \times \ Volume\\ Mass=1\dfrac { kg }{ L } \times 1L\\ Mass =1 kg=1000g \$Step 2: To find the molar mass of waterMolar mass of \$H_2O\$ = \$2 \times H + 1 \times O\$Molar mass of \$H_2O\$ = \$2 \times 1 + 1 \times 16\$Molar mass of \$H_2O\$ = \$ 2 + 16\$ = \$18 \ g/mol\$Step 3: To find the molarity in \$1\$kg of water.Moles in \$ 1 kg \$ of water \$ =\cfrac { mass }{ molar\quad mass } \$ \$ =\cfrac { 1000 \ g }{ 18 \ g/mol } =55.55 \ mol\$Molarity of water = \$ =\cfrac { 55.55\quad moles }{ 1L } =55.55\ M \$Final answer: Option \$C\$Molarity of water is \$55.5M\$

Q: Density of a 2.05 M solution of acetic acid in water is 1.02 g/ml. The molality of the solution is:

Suppose the volume of solution is 1000 ml.\$1.02\ gm\$ in 1 ml of soln.\$1020\ gm\$ in 1000 ml of soln.Number of moles \$=2\cdot 05M\times 1\ lit\$\$=2. 05 \ mol\$Mass of solute \$ =n\times M\cdot wt\$\$=2. 05\times 60=123\ gm\$Mass of solvent \$=1020-123=897\ gm\$Molality \$=\dfrac {123}{60}\times \dfrac {1000}{897}\$\$=2.28\ mol \ kg^{-1}\$

Q: If the density of methanol is \$0.793\; \text{kg} \; { L }^{ -1 }\$, what is its volume needed for making \$2.5\$ L of its \$0.25\$ M solution?

Mass of methanol required is\$\displaystyle M = \text {Volume of solution} \times \text {Molariy of solution} \times \text {Molar mass of methanol}\$\$= 2.5 \times 0.25 \times 32 = 20 \$ g.Volume of methanol needed \$\displaystyle = \frac {\text{mass}}{\text{density}} = \frac {20}{0.793} = 25.22\$ mL.

Question 1

Calculate (a) molality (b) molarity and(c) mole fraction of $KI$ if the density of $20\%$ (mass/ mass) aqueous $KI$ is $1.202\ g\ mL^{-1}$

Accepted Answer

Molality $ \dfrac{w}{w}\% = 20 $ $100\ gm $ solution has $20\ gm $   KI $80\ gm $ solvent has $20\ gm $ KI $ m = \dfrac{\dfrac{20}{166}}{\dfrac{80}{1000}} = 1.51 mol $  Molarity Volume = $\dfrac{m}{d} \Rightarrow \dfrac{20}{1.2} = 83.2 $ Molalrity $ = \dfrac{.120 \times 1000}{83.2} $ $ = 1.44\ m $ Mole fraction   = \dfrac{\dfrac{20}{166}}{\dfrac{80}{18} + \dfrac{20}{166}}

Question 2

The density of a 3M solution of NaCl is 1.25g/ml. Calculate the molality of the solution.

Accepted Answer

3 Molar solution means  there are 3 moles of  NaCl salt in 1 Liter.  Molecular weight of NaCl = 58.44.    Hence, there are $3 \times 58.44$ gms in 1 Litre of water.Density $=\dfrac{ mass}{volume}$Mass of 1 litre of solution = 1.25 gms/ml * 1000 ml = 1, 250 gmsV = volume of water added to make the solution  175.32 gms + 1 gm /ml * V = 1, 250 gm       $\implies V =  1, 074.68 ml  $So 1,074.68 ml or 1, 074.68 gms of water is mixed with 3 moles of Na Cl to make the 3 M solution.            Molality = mass of solute in number of moles / mass of solvent in kg               $= \dfrac{ 3 }{ 1.07468}$kg               $=  2.7915 $ Molal

Question 3

Calculate the concentration of nitric acid in moles per litre in a sample which has a density $1.41 g \displaystyle { mL }^{ -1 }$ and the mass per cent of nitric acid in it being $69\%$.

Accepted Answer

The density  of the solution is 1.41 g/mL.1000 g of the solution will have a volume $\displaystyle = \frac {\text{Mass}}{\text{Density}} = \frac {1000}{1.41}= 709$ ml.The mass per cent of nitric acid is $69\%$.1000 g of the solution will have $\displaystyle 1000 \times \frac {69}{100} = 690 $ g of nitric acid.The molarity of solution is $\displaystyle \frac {\text {Mass of nitric acid}}{\text {Molar mass of nitric acid } \times \text {Volume of solution}}  = \dfrac {690}{63 \times 709} = 0.0155 $ M.

Question 4

Calculate the morality of $NaOH$ in the solution prepared by dissolving its $4\ g$ in water to make solution of $250\ ml$.

Accepted Answer

&#10;&#10;Weight of $NaOH =4 g$&#10;&#10;Molecular weight of $NaOH =40 G$&#10;&#10;Volume of solution is $250ml$&#10;&#10;Molarity = ?&#10;&#10;$M=\frac{W}{MW}\times \frac{1000}{V in ml}$&#10;&#10;$M=\frac{4}{40}\times \frac{1000}{250}$$0.4M$&#10;&#10;&#10;&#10;

Question 5

50 kg of N2(g) and 10kg of $H_2(g)$ are mixed to produce $NH_3 (g)$ . Calculate the amount of $NH_3$ formed. Identify the limiting reagent in the production of the $NH_3$.

Accepted Answer

As we know that,$\text{no. of moles} = \cfrac{\text{Wt.}}{\text{mol. wt.}}$Weight of ${N}_{2} = 50 \; kg$Molecular weight of ${N}_{2} = 28 \; g$No. of moles of ${N}_{2} = \cfrac{50 \times {10}^{3}}{28} = 17.86 \times {10}^{2} \text{ moles}$Weight of ${H}_{2} = 10 \; kg$Molecular weight of ${N}_{2} = 2 \; g$No. of moles of ${N}_{2} = \cfrac{10 \times {10}^{3}}{2} = 5 \times {10}^{3} \text{ moles}$${N}_{2} + 3 {H}_{2} \longrightarrow 2 N{H}_{3}$From the above reaction,$1$ mole of ${N}_{2}$ react with $3$ moles of ${H}_{2}$.No. of moles of ${H}_{2}$ required to react with $17.86 \times {10}^{2}$ moles of ${N}_{2} = 3 \times 17.86 \times {10}^{2} = 5.36 \times {10}^{3} \text{ moles}$But only $5 \times {10}^{3}$ moles of ${H}_{2}$ are available.Thus ${H}_{2}$ is the limiting reagent here.Now, again from the above reaction,Amount of ammonia formed when $3$ moles of ${H}_{2}$ react $= 2 \text{ moles}$Therefore,Amount of ammonia formed when $5 \times {10}^{3}$ moles of ${H}_{2}$ react $= \cfrac{2}{3} \times \left( 5 \times {10}^{3} \right) = 3.33 \times {10}^{3} \text{ moles}$Molecular weight of ammonia $= 17 \; g$Weight of ammonia in $3.33 \times {10}^{3}$ moles $= 17 \times \left( 3.33 \times {10}^{3} \right) = 56.61 \; kg$.Hence the amount of $N{H}_{3}$ formed is $56.61 \; kg$.

Question 6

The molarity of pure water is:

Accepted Answer

Formula:$Molarity=\cfrac { No\ of\ moles\ of\ solute }{ Volume\ of\ solution\ in\ litre } $Step 1: To find the mass of water in one liter.We know that,Density of water = $1 g/ml$             or = $ 1 kg/L$Therefore, the mass of $ 1 L $ of water will be:$ Mass =\ Density\ of\ water\ \times \ Volume\ Mass=1\dfrac { kg }{ L } \times 1L\ Mass =1 kg=1000g $Step 2: To find the molar mass of waterMolar mass of $H_2O$ = $2 \times H + 1 \times O$Molar mass of $H_2O$ = $2 \times 1 + 1 \times 16$Molar mass of $H_2O$ = $ 2 + 16$ = $18 \ g/mol$Step 3: To find the molarity in $1$kg of water.Moles in $ 1 kg $ of water $ =\cfrac { mass }{ molar\quad mass } $ $ =\cfrac { 1000 \ g }{ 18 \ g/mol } =55.55 \ mol$Molarity of water = $ =\cfrac { 55.55\quad moles }{ 1L } =55.55\ M $Final answer: Option $C$Molarity of water is $55.5M$

Question 7

Accepted Answer

Question 8

Calculate the molarity of a solution of ethanol in water in which the mole fraction of ethanol is $0.040$.[Assume the density of water to be one]

Accepted Answer

The mole fraction of ethanol is equal to the ratio of the number of moles of ethanol to the total number of moles present in the solution.For dilute solutions, 1 L of water (or 1 L of solution) contains $\displaystyle \frac {1000}{18} = 55.55 $ moles of water. Let n be the number of moles of ethanol. The mole fraction of ethanol is 0.040.Hence, $\displaystyle 0.040 = \frac {n}{n+55.55} $$\displaystyle n = 0.040n +  2.222$$\displaystyle n = \frac {2.222}{1-0.040} = 2.31 $ moles.The molarity of the solution is the ratio of the number of moles of solute present in 1 L of solution.As 1 L of solution contains 2.31 moles of ethanol, the molarity of a solution is 2.31 M.

Question 9

Density of a 2.05 M solution of acetic acid in water is 1.02 g/ml. The molality of the solution is:

Accepted Answer

Suppose the volume of solution is 1000 ml.$1.02\ gm$ in 1 ml of soln.$1020\ gm$ in 1000 ml of soln.Number of moles $=2\cdot 05M\times 1\ lit$$=2. 05 \ mol$Mass of solute $ =n\times M\cdot wt$$=2. 05\times 60=123\ gm$Mass of solvent $=1020-123=897\ gm$Molality $=\dfrac {123}{60}\times \dfrac {1000}{897}$$=2.28\ mol \ kg^{-1}$

Question 10

If the density of methanol is $0.793\; \text{kg} \; { L }^{ -1 }$, what is its volume needed for making $2.5$ L of its $0.25$ M solution?

Accepted Answer

Mass of methanol required is$\displaystyle M = \text {Volume of solution} \times \text {Molariy of solution} \times \text {Molar mass of methanol}$$= 2.5 \times 0.25 \times 32 = 20 $ g.Volume of methanol needed $\displaystyle = \frac {\text{mass}}{\text{density}} = \frac {20}{0.793} = 25.22$ mL.

Revise with Concepts

Concentration Terms: Molarity

Concentration Terms: Molality

Quick Summary With Stories

Concentration Term: Molarity

Concentration Term: Molality

Calculate
(a) molality
(b) molarity and
(c) mole fraction of $K I$ if the density of $20 %$ (mass/ mass) aqueous $K I$ is $1.202 g m L^{- 1}$

The density of a 3M solution of NaCl is 1.25g/ml. Calculate the molality of the solution.

Calculate the concentration of nitric acid in moles per litre in a sample which has a density $1.41 g m L^{- 1}$ and the mass per cent of nitric acid in it being $69 %$ .

Calculate the morality of $N a O H$ in the solution prepared by dissolving its $4 g$ in water to make solution of $250 m l$ .

50 kg of N2(g) and 10kg of $H_{2} (g)$ are mixed to produce $N H_{3} (g)$ . Calculate the amount of $N H_{3}$ formed. Identify the limiting reagent in the production of the $N H_{3}$ .

The molarity of pure water is:

Calculate the molarity of a solution of ethanol in water in which the mole fraction of ethanol is $0.040$ .

[Assume the density of water to be one]

Density of a 2.05 M solution of acetic acid in water is 1.02 g/ml. The molality of the solution is:

If the density of methanol is $0.793 kg L^{- 1}$ , what is its volume needed for making $2.5$ L of its $0.25$ M solution?

Revise with Concepts

Concentration Terms: Molarity

Concentration Terms: Molality

Quick Summary With Stories

Concentration Term: Molarity

Concentration Term: Molality

Calculate (a) molality (b) molarity and (c) mole fraction of KI if the density of 20% (mass/ mass) aqueous KI is 1.202 g mL−1

The density of a 3M solution of NaCl is 1.25g/ml. Calculate the molality of the solution.

Calculate the concentration of nitric acid in moles per litre in a sample which has a density 1.41gmL−1 and the mass per cent of nitric acid in it being 69%.

Calculate the morality of NaOH in the solution prepared by dissolving its 4 g in water to make solution of 250 ml.

50 kg of N2(g) and 10kg of H2​(g) are mixed to produce NH3​(g) . Calculate the amount of NH3​ formed. Identify the limiting reagent in the production of the NH3​.

The molarity of pure water is:

Calculate the molarity of a solution of ethanol in water in which the mole fraction of ethanol is 0.040. [Assume the density of water to be one]

Density of a 2.05 M solution of acetic acid in water is 1.02 g/ml. The molality of the solution is:

If the density of methanol is 0.793kgL−1, what is its volume needed for making 2.5 L of its 0.25 M solution?

Calculate
(a) molality
(b) molarity and
(c) mole fraction of $K I$ if the density of $20 %$ (mass/ mass) aqueous $K I$ is $1.202 g m L^{- 1}$

Calculate the concentration of nitric acid in moles per litre in a sample which has a density $1.41 g m L^{- 1}$ and the mass per cent of nitric acid in it being $69 %$ .

Calculate the morality of $N a O H$ in the solution prepared by dissolving its $4 g$ in water to make solution of $250 m l$ .

50 kg of N2(g) and 10kg of $H_{2} (g)$ are mixed to produce $N H_{3} (g)$ . Calculate the amount of $N H_{3}$ formed. Identify the limiting reagent in the production of the $N H_{3}$ .

Calculate the molarity of a solution of ethanol in water in which the mole fraction of ethanol is $0.040$ .

[Assume the density of water to be one]

If the density of methanol is $0.793 kg L^{- 1}$ , what is its volume needed for making $2.5$ L of its $0.25$ M solution?