When mL methanol (molar mass = 32 g; density = 0.792 ) is added to 100 mL water (density = 1 ), the following diagram is obtained.

Question

When $‘ x ’ \times 10^{- 2}$ mL methanol (molar mass = 32 g; density = 0.792 $g / {cm}^{3}$ ) is added to 100 mL water (density = 1 $g / {cm}^{3}$ ), the following diagram is obtained.

x = ___________ (nearest integer)

[Given: Molal freezing point depression constant of water at 273.15 K is 1.86 K kg ${mol}^{- 1}$ ] [2024]

Smart Achievers · Accepted Answer

(543)

$Δ T_{f} = m k_{f}$

$(273.15 - 270.65) K = m \times 1.86 {K kg mol}^{- 1}$

$m = \frac{2.5}{1.86} m$

$\frac{W_{methanol} \times 1000}{M_{methanol} \times W_{water} (in g)} = \frac{2.5}{1.86}$

$\frac{(d_{methanol} \times V_{methanol}) \times 1000}{M_{methanol} \times (d_{water} \times V_{water})} = \frac{2.5}{1.86}$

$\frac{(0.792 {g m L}^{- 1} \times x \times 10^{- 2} mL) \times 1000}{32 {g mol}^{- 1} \times (1 {gm L}^{- 1} \times 100 mL)} = \frac{2.5}{1.86} {mol kg}^{- 1}$

$x = 543$

Solution

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Solution

Q. When ‘x’×10-2 mL methanol (molar mass = 32 g; density = 0.792 g/cm3) is added to 100 mL water (density = 1 g/cm3), the following diagram is obtained. x = ___________ (nearest integer) [Given: Molal freezing point depression constant of water at 273.15 K is 1.86 K kg mol-1] [2024]

Ans. (543) ΔTf=mkf (273.15-270.65)K=m×1.86 K kg mol-1 m=2.51.86m Wmethanol×1000Mmethanol×Wwater (in g)=2.51.86 (dmethanol×Vmethanol)×1000Mmethanol×(dwater×Vwater)=2.51.86 (0.792 g m L-1×x×10-2 mL)×100032 g mol-1×(1 gm L-1×100 mL)=2.51.86 mol kg-1 x=543

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