Solution - as we know that, âˆ†Tf = Kf × m m = ΔTFKf =0·301 86 = 0 161 According to Rault's law, P° - pp0 = nBnA ( here, nB and nA are no Of moles of solute and solvent respectively ) ; 23 51 - p23·51 =0·1611000/18 = 0 0020898 : P = 23 51-23 51×0 0020989 P= 23 46mmHg Which is near to 23 44 in all options So, option 3 is correct

explain in detail

Solution -
as we know that,
∆T_f = K_f × m

m =

\frac{Δ T_{F}}{K_{f}}

\frac{0 \cdot 30}{1.86}

= 0.161
According to Rault's law,
P° -

\frac{p}{p^{0}}

\frac{n_{B}}{n_{A}}

( here, n_Band n_A are no. Of moles of solute and solvent respectively.)

; 23.51 -

\frac{p}{23 \cdot 51}

\frac{0 \cdot 161}{1000 / 18}

= 0.0020898
: P = 23.51-23.51×0.0020989
P= 23.46mmHg
Which is near to 23.44 in all options . So,
option 3 is correct.

View Full Answer