The freezing point of a solution prepared from 1.25 g of non-electrolyte and 20 g of water is 271.9K. If molar depression constant is 1 86 K m-1 then molar mass of the solute will be

Question

The freezing point of a solution prepared from 1.25 g of non-electrolyte and 20 g of water is 271.9K. If molar depression constant is 1 86 K m-1 then molar mass of the solute will be (A) 105.7 (B) 106.7 (C) 115.3 (D) 93

Tardigrade Admin · Accepted Answer

Δ T f =(1000 K f × W 2/ Mw 2 × W 1) [Δ T =273-271.9=1.25] 1.25=(1000 × 1.86 × 1.25/ Mw 2 × 20) Mw 2=93

Q. The freezing point of a solution prepared from 1.25 g of non-electrolyte and 20 g of water is 271.9K. If molar depression constant is 1 86 $K m^{- 1}$ then molar mass of the solute will be

105.7

106.7

115.3

93

$Δ T_{f} = \frac{1000 K _{f} \times W _{2}}{M w _{2} \times W _{1}}$
[\Delta T =273-271.9=1.25] $< b r / > 1.25 = \frac{1000 \times 1.86 \times 1.25}{M w _{2} \times 20}$
$M w_{2} = 93$

Q. The freezing point of a solution prepared from 1.25 g of non-electrolyte and 20 g of water is 271.9K. If molar depression constant is 1 86 Km−1 then molar mass of the solute will be

105.7

106.7

115.3

93

ΔTf​=Mw2​×W1​1000Kf​×W2​​ [\Delta T =273-271.9=1.25] <br/>1.25=Mw2​×201000×1.86×1.25​ Mw2​=93

Q. The freezing point of a solution prepared from 1.25 g of non-electrolyte and 20 g of water is 271.9K. If molar depression constant is 1 86 $K m^{- 1}$ then molar mass of the solute will be

$Δ T_{f} = \frac{1000 K _{f} \times W _{2}}{M w _{2} \times W _{1}}$
[\Delta T =273-271.9=1.25] $< b r / > 1.25 = \frac{1000 \times 1.86 \times 1.25}{M w _{2} \times 20}$
$M w_{2} = 93$