Project ForgeSign up free
Process Calculations
Material Balance
GATE 2010
Hard

A saturated solution at 30°C contains 5 moles of solute (M.W.=50kg/kmol) per kg of solvent (M.W.=20kg/kmol). The solubility at 100°C is 10 moles of the solute per kg of the solvent. If 10kg of the original solution is heated to 100°C, then the weight of the additional solute that can be dissolved in it, is

  • A0.25 kg
  • B1 kg
  • 2 kg
  • D3.34 kg

Explanation

Original solution: 5mol×50kg/kmol=250g solute per 1000g solvent, i.e. 1250g total solution is 20% solute, 80% solvent. 10kg (10000g) of this solution contains 2000g (2kg) solute and 8000g (8kg) solvent. At 100°C, solubility=10mol×50=500g solute per kg solvent, so 8kg solvent can hold 8×0.5=4kg solute total. Additional solute dissolvable=4−2=2kg.

Source: GATE CH 2010 (2010)

This question is part of a 15-day revision cycle — sign in and Project Forge will bring it back the day before you'd forget it.

Start revising free