Q: Calculate the mole fraction of benzene in solution containing \$30\$% by mass in \$CCl_4\$?

Let the mass of solution be \$100 \ gm\$Mass of Benzene \$= 30 \ gm\$Mass of \$CCl_4=70 \ gm\$Moles of Benzene \$= \cfrac {30}{78}=0.385\$Moles of \$CCl_4=\cfrac {70}{154}=0.45 \ moles\$Mole fraction (Benzene) \$=\cfrac {0.385}{0.385+0.45}=0.461\$

Q: Henry's Law constant for \$ CO_2\$ in water is \$1.67 \times 10^8\$ Pa at 298 K. Calculate the quantity in 1 L of soda water when packed under 2.5 atm pressure at 298 K.

We know Henry's Law: \$P=K_{H}\times x\$ \$\Rightarrow x=\dfrac{2.5\times 1.013\times 10^{5}}{1.6\times 10^{8}}\$ \$\Rightarrow \dfrac{n_{CO_{2}}}{1000/18}=\dfrac{2.5\times 1.013}{1.6\times 1000}\$ \$\Rightarrow n_{CO_{2}}=\dfrac{25}{16}\times \dfrac{1.013}{18}\$ \$\Rightarrow \$ mass of \$CO_{2}\$ dissolved \$=\dfrac{25}{16}\times \dfrac{44\times 1.013}{18}\$ \$=3.86\ grams.\$

Q: The vapour pressure of pure liquids \$A\$ and \$B\$ are \$450\$ and \$700\$ mm Hg respectively. at \$350\$ K. Find out the composition of the liquid mixture if total vapour pressure is \$600\$ mm Hg. Also, find the composition of the vapour phase.

Given: Vapour pressure of pure liquid A, \$PoA=450mmofHg\$Vapour pressure of pure liquid A, \$PoA=700mm\quad of\quad Hg\$Total vapour pressure,a \$Ptotal=600mmofHg\$Use the formula of Raoult’s law\$\$600=(450–700)XA+700\$\$250XA=100\$\$XA=\$\$\dfrac { 100 }{ 250 } \$ \$=0.4\$Use formula \$XB=1-XA\$Substitute the valueswe get, \$XB=1−0.4=0.6\$use formula \$PA=PoA×XA=450×0.4=180mm\quad of\quad Hg\$\$PB=PoB×XB=700×0.6=420mm\quad of\quad Hg\$Now, in the vapour phase:Mole fraction of liquid \$A=\dfrac { 180 }{ 180+420 } =0.30\$Mole fraction of liquid \$B,YB=1−YA=1–0.30=0.70

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: 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: Calculate molality of 2.5 grams of Ethanoic Acid \$(CH_3COOH)\$ in 75 grams of Benzene.

The molar mass of ethanoic acid is \$\displaystyle 2(12)+2(16)+4(1)=60\$ g/mol. The mass of ethanoic acid is divided with molar mass of ethanoic acid to obtain number of moles of ethanoic acid.Number of moles of ethanoic acid \$\displaystyle = \dfrac {2.5 \: g}{60 \: g/mol}=0.04167 \: mol\$The mass of benzene is 75 grams or \$\displaystyle \dfrac {75}{1000}=0.075\$ kg.The molality of ethanoic acid in benzene is the ratio of the number of moles of ethanoic acid to the mass of benzene (in kg). It is \$\displaystyle \dfrac {0.04167 \: mol}{0.075 \: kg}=0.556 \: m\$

Question 1

Calculate the mole fraction of benzene in solution containing $30$% by mass in $CCl_4$?

Accepted Answer

Let the mass of solution be $100 \ gm$Mass of Benzene $= 30 \ gm$Mass of $CCl_4=70 \ gm$Moles of Benzene $= \cfrac {30}{78}=0.385$Moles of $CCl_4=\cfrac {70}{154}=0.45 \ moles$Mole fraction (Benzene) $=\cfrac {0.385}{0.385+0.45}=0.461$

Question 2

Henry's Law constant for $ CO_2$ in water is $1.67 \times 10^8$ Pa at 298 K. Calculate the quantity in 1 L of soda water when packed under 2.5 atm pressure at 298 K.

Accepted Answer

We know Henry's Law: $P=K_{H}\times x$ $\Rightarrow x=\dfrac{2.5\times 1.013\times 10^{5}}{1.6\times 10^{8}}$ $\Rightarrow \dfrac{n_{CO_{2}}}{1000/18}=\dfrac{2.5\times 1.013}{1.6\times 1000}$ $\Rightarrow n_{CO_{2}}=\dfrac{25}{16}\times \dfrac{1.013}{18}$ $\Rightarrow $ mass of $CO_{2}$ dissolved $=\dfrac{25}{16}\times \dfrac{44\times 1.013}{18}$ $=3.86\ grams.$

Question 3

The vapour pressure of pure liquids $A$ and $B$ are $450$ and $700$ mm Hg respectively. at $350$ K. Find out the composition of the liquid mixture if total vapour pressure is $600$ mm Hg. Also, find the composition of the vapour phase.

Accepted Answer

Given: Vapour pressure of pure liquid A, $PoA=450mmofHg$Vapour pressure of pure liquid A, $PoA=700mm\quad of\quad Hg$Total vapour pressure,a $Ptotal=600mmofHg$Use the formula of Raoult&#8217;s law600=(450&#8211;700)XA+700250XA=100XA=\dfrac { 100 }{ 250 } =0.4$Use formula $XB=1-XA$Substitute the valueswe get, $XB=1&#8722;0.4=0.6$use formula $PA=PoA&#215;XA=450&#215;0.4=180mm\quad of\quad HgPB=PoB&#215;XB=700&#215;0.6=420mm\quad of\quad Hg$Now, in the vapour phase:Mole fraction of liquid $A=\dfrac { 180 }{ 180+420 } =0.30$Mole fraction of liquid $B,YB=1&#8722;YA=1&#8211;0.30=0.70

Question 4

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 5

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 6

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 7

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 8

Calculate molality of 2.5 grams of Ethanoic Acid $(CH_3COOH)$ in 75 grams of Benzene.

Accepted Answer

The molar mass of ethanoic acid is $\displaystyle 2(12)+2(16)+4(1)=60$ g/mol. The mass of ethanoic acid is divided with molar mass of ethanoic acid to obtain number of moles of ethanoic acid.Number of moles of ethanoic acid $\displaystyle = \dfrac {2.5 \: g}{60 \: g/mol}=0.04167 \: mol$The mass of benzene is 75 grams or $\displaystyle \dfrac {75}{1000}=0.075$ kg.The molality of ethanoic acid in benzene is the ratio of the number of moles of ethanoic acid to the mass of benzene (in kg). It is $\displaystyle \dfrac {0.04167 \: mol}{0.075 \: kg}=0.556 \: m$

Question 9

40 grams of sodium hydroxide is dissolved in enough water to make $1$ liter of solution. Determine the molarity of the solution?

Accepted Answer

The molar mass of NaOH is 40 g/mol.The number of moles of NaOH is equal to the ratio of its mass to molar mass.The number of moles of NaOH $\displaystyle  = \dfrac {40 g}{40 g/mol} = 1 mol$ The molarity of NaOH solution is the ratio of number of moles of NaOH to total volume of solution in L.$\displaystyle  M = \dfrac {1mol}{1L}=1 M$

Question 10

$H_{2}S$ a toxic gas with rotten egg like smell, is used for the qualitative analysis. If the solubility of $H_{2}S$ in water at $STP$ is $0.195\ m$, calculate Henry's law constant.

Accepted Answer

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