A solution containing 10 g of a non-volatile, higher nonelectrolyte and 400 g of water boils at 100 .256o C at 1 atm . The molecular weight of the solute (in g/mol) is Given: ( Kb for water 0.512 οC / m )

Question

A solution containing 10 g of a non-volatile, higher nonelectrolyte and 400 g of water boils at 100 .256o C at 1 atm . The molecular weight of the solute (in g/mol) is Given: ( Kb for water 0.512 οC / m ) (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

Δ Tb=i × Kb × m 0.256=1 × 0.512 ×(((10.0/M)/0.4)) M=(0.512 × 10.0/0.4 × 0.256) M =50 g / mol

Q. A solution containing $10 g$ of a non-volatile, higher nonelectrolyte and $400 g$ of water boils at $100.25 6^{o} C$ at $1 a t m$ . The molecular weight of the solute $(in g / m o l)$ is
Given: ( $K_{b}$ for water $0.512^{ο} C / m$ )

$Δ T_{b} = i \times K_{b} \times m$
$0.256 = 1 \times 0.512 \times (\frac{\frac{10.0}{M}}{0.4})$
$M = \frac{0.512 \times 10.0}{0.4 \times 0.256}$
$M = 50 g / m o l$

Q. A solution containing 10g of a non-volatile, higher nonelectrolyte and 400g of water boils at 100.256oC at 1atm . The molecular weight of the solute (ing/mol) is Given: ( Kb​ for water 0.512οC/m )

ΔTb​=i×Kb​×m 0.256=1×0.512×(0.4M10.0​​) M=0.4×0.2560.512×10.0​ M=50g/mol

Q. A solution containing $10 g$ of a non-volatile, higher nonelectrolyte and $400 g$ of water boils at $100.25 6^{o} C$ at $1 a t m$ . The molecular weight of the solute $(in g / m o l)$ is
Given: ( $K_{b}$ for water $0.512^{ο} C / m$ )

$Δ T_{b} = i \times K_{b} \times m$
$0.256 = 1 \times 0.512 \times (\frac{\frac{10.0}{M}}{0.4})$
$M = \frac{0.512 \times 10.0}{0.4 \times 0.256}$
$M = 50 g / m o l$