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\%\$.

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.

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.

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>> $Concentration Terms: Molality$

Concentration Terms: Molality

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

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

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: 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]

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.

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}\$

Chemistry

definition

Molality

Molality(m) of a solution is defined as the number of moles of solute dissolved in 1000g of a solvent.It does not depend on temperature. Mathematically, it is expressed as,

\frac{N u m b e r o f m o l e s o f s o l u t e}{W e i g h t o f s o l v e n t i n k g}

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Concentration Terms: Molarity

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Important Questions

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

Medium

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

Medium

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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 $69 %$ .

Medium

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

Hard

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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 $N H_{3}$ .

Medium

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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]

Medium

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

Hard

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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?

Medium

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definition

Molality

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Concentration Terms: Molarity

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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?