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

# 1.20g sample of Na2CO3 and K2CO3 was dissolved in water to form 100ml of a solution. 20ml of this solution required 40ml of 0.1N HCl for complete neutralization. Calculate the weight of Na2CO3 in the mixture.

Open in App
Solution

## Total weight of the mixture(Na2CO3+K2CO3)=1.20gVolume of the solution=100mlVolume of solution taken for neutralisation V2= 20 mlVolume of HCl solution V1=40 mlNormality of HCl, N1=0.1 NTo calculate the mass of Na2CO3N1×V1=N2×V20.1×40=N2×20N2=0.2N1 l of solution contain no. of g equivalent=0.20.1 l of solution contain no. of g equivalent=0.1*0.2=0.02 g equivalentNo. of g equivalent=sum of (g equivalent of Na2CO3+g equivalent of K2CO3)No. of g equivalent=mass/equivalent massequivalent mass=molar mass/valenceMolar mass of Na2Co3=106 gMolar mass of K2CO3=138 gEquivalent mass of Na2CO3=106/2=53 gEquivalent mass of K2CO3=138/2=69gConsider mass of Na2CO3=x gMass of K2CO3=(1.2-x)gSo we can write0.02=x/53+1.2-x/69x=0.59 g=0.6g

Suggest Corrections
0
Join BYJU'S Learning Program
Related Videos
Percentage Composition and Molecular Formula
CHEMISTRY
Watch in App
Explore more
Join BYJU'S Learning Program