1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

# 100 g of a sample of common salt containing contamination of NH4Cl (Molar mass = 53.5 g/mol) and MgCl2 (Molar mass = 95 g/mol) to the extent of 2% each by mass is dissolved in water. Which of the following is/are correct when volume of 5% by mass of AgNO3 (Molar mass = 170 g/mol) solution (d = 1.04 g/cm3) is required to precipitate all chloride ions?

A
Molarity of AgNO3 solution = 0.3 M
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
Volume of AgNO3 required to precipiate all chloride ions = 570 L
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
Molarity of AgNO3 solution = 0.03 M
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
Volume of AgNO3 required to precipiate all chloride ions = 5.7 L
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
Open in App
Solution

## The correct option is D Volume of AgNO3 required to precipiate all chloride ions = 5.7 LAmount of NH4Cl=2 g=253.5=0.037 mol Cl− ions in NH4Cl=0.037 mol Amount of MgCl2=2 g=295=0.021 mol Cl− ions in MgCl2=2×0.021=0.042 mol Amount of NaCl=96 g=9658.5=1.63 mol Total Cl−ions=0.037+0.042+1.63=1.71 moles AgNO3+Cl−→ AgCl+NO−3 From the reaction we can see that , Moles of AgNO3 required for precipitation=1.71 mol Molarity of AgNO3 =%× d×10Molar mass=5×1.04×10170=0.3 M ∴ Volume of AgNO3=1.710.3=5.7 L

Suggest Corrections
0
Join BYJU'S Learning Program
Explore more
Join BYJU'S Learning Program