Solutions MCQs for NEET — Chemistry Questions with Answers

Molarity of 95% $H_{2}S{O}_4=\frac{95}{98}\times \frac{1}{100/1.85}\times 1000$ = 17.93 M

Molarity of 15% $H_{2}S{O}_4=\frac{15}{98}\times \frac{1}{100/1.10}\times 1000$ = 1.68 M

$M_1{V}_1$ = $M_2{V}_2$

(95% H₂SO₄) = (15% H₂SO₄) 

$17.93\times V_1=1.68\times 100$ or V₁ = 9.4 cm³

$N=M\times \frac{\text{molecular mass (}{M}_{\text{2}}\text{)}}{\text{Equivalent mass (}E\text{)}}$

So, $M=N\times \frac{\text{Equivalent mass (}E\text{)}}{\text{Molecular mass (}{M}_{\text{2}})}$

Equivalent mass of salt $=\frac{\text{Molecular mass}}{\text{Total positive valency}}$

Equivalent mass of $N{a}_{2}C{O}_3=\frac{M_2}{2}$

$M=0.2\times \frac{M_2/2}{M_2}$; $M=\frac{0.2}{2}$ = 0.1 M 

H₂SO₄ is 95% by volume

Wt. of H₂SO₄ = 95 g

Vol. of solution = 100 ml

∴ moles of $H_{2}S{O}_4=\frac{95}{98}$ and weight of solution $=100\times 1.98=198g$

Weight of water $=198-95=103g$

Molality $=\frac{95\times 1000}{98\times 103}=9.412$

Hence molality of H₂SO₄ solution is 9.412  

Given H₂SO₄ is 93% by volume

Wt. of H₂SO₄ = 93 g

Volume of solution = 100 ml ∵ Density $=\frac{\text{mass}}{\text{volume}}$ 

$\therefore mass=d\times$volume

∴ weight of solution $=100\times 1.84g=184g$

wt. of water $=184-93=91g$

Molality $=\frac{\text{Moles}}{\text{wt. of water in }kg}=\frac{93\times 1000}{98\times 91}=10.42$

Mass of methanol = 16 g, Mol. mass of CH₃OH = 32

∴ No. of moles of methanol $=\frac{16}{32}=0.5moles$

No. of moles of water = $\frac{90}{18}=5moles$

∴ Mole fraction of methanol $=\frac{0.5}{5+0.5}=0.090$   

Since 18 g of water = 1mole

25 g of water = $\frac{25}{18}=1.38$ mole

Similarly, 46 g of ethanol = 1 mole

25 g of ethanol $=\frac{25}{46}=0.55$moles

Again, 60 g of acetic acid = 1 mole

50 g of acetic acid $=\frac{50}{60}=0.83$mole

∴ Mole fraction of water $=\frac{1.38}{1.38+0.55+0.83}=0.50$

Similarly, Mole fraction of ethanol $=\frac{0.55}{1.38+0.55+0.83}=0.19$  

Mole fraction of acetic acid $=\frac{0.83}{1.38+0.55+0.83}=0.3$

$\frac{P_A^0-P_A}{P_A^0}=\frac{W_B/M_A}{W_B/M_B+W_A/M_A}=\frac{34.2/342}{34.2/342+180/18}=\frac{0.1}{10.1}=0.0099$   

$\frac{P_A^0-P_A}{P_A^0}=\text{Molality}\times (1-\alpha x+x\alpha +\gamma x)$

The value of $P_A^0-P_A$ is maximum for BaCl₂.

$\frac{P^0-P_s}{P^0}=\frac{n_2}{n_1}$; $\frac{143-P_s}{143}=\frac{0.5/65}{158/154}$

or $P_s=141.93mm$  

For equimolar solutions, $X_B=X_T=0.5$

$P_B=X_B\times P_B^0=0.5\times 160=80mm$

$P_T=X_T\times P_T^0=0.5\times 60=30mm$

$P_{\text{Total}}=80+30=110mm$

Mole fraction of toluene in vapour phase $=\frac{30}{110}=0.27$