At solution containing 12.5 g of non-electrolyte substance in 175 g of water gave boiling point elevation of 0.70 K. Calculate the molar mass of the…

Question

At solution containing 12.5 g of non-electrolyte substance in 175 g of water gave boiling point elevation of 0.70 K. Calculate the molar mass of the substance. Molal elevation constant (K_b) for water 0.52 K kg mol^(-1) ?

MARKS Content Team · Accepted Answer

Correct answer: (A) 53. ΔT_b=k_b×w/M×1000/W M=(K_b×w×1000)/(ΔT_b×W) M_B=(K_b×W_B×1000)/(ΔT_b×W_A) According to available data Mass of solute W_B=12.5 g Mass of solute W_A=175 g Molal elevation constant K_b=0.52 K kg mol^(-1) Elevation in boiling point temperature ΔT_b=0.70 K ∴M=(0.52 K kg mol^(-1)×12.5 g×1000)/(0.70 K×175 g) =53 g mol^(-1)

At solution containing 12.5 g of non-electrolyte substance in 175 g of water gave boiling point elevation of 0.70 K. Calculate the molar mass of the substance. Molal elevation constant $(K_{b})$ for water $0.52 K k g m o l^{- 1}$ ?

Solution

More Solutions Questions

At solution containing 12.5 g of non-electrolyte substance in 175 g of water gave boiling point elevation of 0.70 K. Calculate the molar mass of the substance. Molal elevation constant (Kb) for water 0.52 K kg mol-1 ?

Solution

More Solutions Questions

At solution containing 12.5 g of non-electrolyte substance in 175 g of water gave boiling point elevation of 0.70 K. Calculate the molar mass of the substance. Molal elevation constant $(K_{b})$ for water $0.52 K k g m o l^{- 1}$ ?