100 mL solution of NaOH (containing 4 g NaOH per litre) and 50 mL of HCl (containing 7.3 g HCl per litre) react as follows, NaOH(aq)+HCl(aq)⟶NaCl(aq)+H2O(l) 0.5 g of NaCl is formed. Thus, unreacted NaOH is:
A
0.058 g
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B
3.66 g
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C
10.8 g
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D
0.63 g
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Solution
The correct option is A0.058 g NaOH(aq)+HCl(aq)⟶NaCl(aq)+H2O(l) 0.4 g 0.365 g 0.01 mol 0.01 mol So, according to this, 0.585 g of NaCl should be formed but, only 0.5 g forms. Remaining NaOH is 0.4×(0.585−0.5)0.585=0.058 g.