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Question

Chemistry Question on Stoichiometry and Stoichiometric Calculations

0.45g0.45 \,g of acid molecular weight 9090 is neutralised by 20ml20\, ml of 0.5N0.5\,N caustic potash. The basicity of acid is

A

1

B

2

C

3

D

4

Answer

2

Explanation

Solution

Let the nfactor n _{\text {factor }} be nn.
Moles of acid =0.4590=5=\frac{0.45}{90}=5 milimoles
Comparing equivalents of acid and NaOHNaOH
5×n=0.5×205 \times n=0.5 \times 20
n=2n =2
For acid, nfactor =n _{\text {factor }}= basicity
\therefore Basicity is 22