The freezing point of a solution prepared from 1.25 g of non-electrolyte and 20 g of water is 271.9K. If the molar depression constant is $1.86 K {m o l}^{- 1}$ , then molar mass of the solute will be:

105.7

No worries! We‘ve got your back. Try BYJU‘S free classes today!

106.7

No worries! We‘ve got your back. Try BYJU‘S free classes today!

115.3

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is D 93
$Δ T_{f} = \frac{1000 K_{f} \times W_{2}}{M w_{2} \times W_{1}}$

$[Δ T = 273 - 271.9 = 1.25]$

$1.25 = \frac{1000 \times 1.86 \times 1.25}{M w_{2} \times 20}$

$M w_{2} = 93$

Hence, the correct option is $D$

Suggest Corrections

Similar questions

The freezing point of a solution prepared from 1.25 gm of a non-electrolyte and 20gm of water is 271.9 K. If molar depression constant is 1.86 K mole^-1, then molar mass of the solute will be

Q. 1.00 g of a non-electrolyte solute (molar mass 250g mol

^{- 1}

) was dissolved in 51.2 g of benzene.If the freezing point depression constant K

_{f}

of benzene is 5.12 K kg mol

^{1}

, the freezing point of benzene will be lowered by:

Q. 1 g of non-electrolyte solute dissolved in 50g of benzene lowered the freezing point of benzene by 0.40K. The freezing point depression constant of benzene is 5.12 K kg/mol. Find the molar mass of the solute.

1.00

g of non-electrolyte solute dissolved in

50

g of benzene lowered the point of benzene by

0.40

K. The freezing point depression constant of benzene is

5.12

K kg/mol. Find the molar mass of the solute.

Join BYJU'S Learning Program

The freezing point of a solution prepared from 1.25 g of non-electrolyte and 20 g of water is 271.9K. If the molar depression constant is 1.86Kmol−1, then molar mass of the solute will be:

The correct option is D 93ΔTf=1000Kf×W2Mw2×W1[ΔT=273−271.9=1.25]1.25=1000×1.86×1.25Mw2×20Mw2=93Hence, the correct option is D

The freezing point of a solution prepared from 1.25 g of non-electrolyte and 20 g of water is 271.9K. If the molar depression constant is $1.86 K {m o l}^{- 1}$ , then molar mass of the solute will be:

The correct option is D 93
$Δ T_{f} = \frac{1000 K_{f} \times W_{2}}{M w_{2} \times W_{1}}$

$[Δ T = 273 - 271.9 = 1.25]$

$1.25 = \frac{1000 \times 1.86 \times 1.25}{M w_{2} \times 20}$

$M w_{2} = 93$

Hence, the correct option is $D$