A solution containing 3.24 g of a nonvolatile nonelectrolyte and 200 g of water boils at 100.130° C at 1 atm. What is the molecular weight (in g / mol ) of the solute? ( K b . for water .0.513° C / m )

Question

A solution containing 3.24 g of a nonvolatile nonelectrolyte and 200 g of water boils at 100.130° C at 1 atm. What is the molecular weight (in g / mol ) of the solute? ( K b . for water .0.513° C / m ) (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

Solute - B, Water - A Δ T b =100.130-100 =0.130 Δ T b = K b × ( w B × 1000/ m B × w A ( gm )) 0.130=0.513 × (3.24 × 1000/ m B × 200) m B =63.92 gm mol -1

Q. A solution containing $3.24 g$ of a nonvolatile nonelectrolyte and $200 g$ of water boils at $100.13 0^{\circ} C$ at $1 a t m$ . What is the molecular weight (in $g / m o l$ ) of the solute? $(K_{b}$ for water $0.51 3^{\circ} C / m)$

Solute - $B$ , Water - $A$
$Δ T_{b} = 100.130 - 100$
$= 0.130$
$Δ T_{b} = K_{b} \times \frac{w _{B} \times 1000}{m _{B} \times w _{A} ( g m )}$
$0.130 = 0.513 \times \frac{3.24 \times 1000}{m _{B} \times 200}$
$m_{B} = 63.92 g m m o l^{- 1}$

Q. A solution containing 3.24g of a nonvolatile nonelectrolyte and 200g of water boils at 100.130∘C at 1atm. What is the molecular weight (in g/mol ) of the solute? (Kb​ for water 0.513∘C/m)

Solute - B, Water - A ΔTb​=100.130−100 =0.130 ΔTb​=Kb​×mB​×wA​(gm)wB​×1000​ 0.130=0.513×mB​×2003.24×1000​ mB​=63.92gmmol−1

Q. A solution containing $3.24 g$ of a nonvolatile nonelectrolyte and $200 g$ of water boils at $100.13 0^{\circ} C$ at $1 a t m$ . What is the molecular weight (in $g / m o l$ ) of the solute? $(K_{b}$ for water $0.51 3^{\circ} C / m)$

Solute - $B$ , Water - $A$
$Δ T_{b} = 100.130 - 100$
$= 0.130$
$Δ T_{b} = K_{b} \times \frac{w _{B} \times 1000}{m _{B} \times w _{A} ( g m )}$
$0.130 = 0.513 \times \frac{3.24 \times 1000}{m _{B} \times 200}$
$m_{B} = 63.92 g m m o l^{- 1}$