A 20.0 mL solution of Na 2 CO 3 required 30 mL of 0.01 M K 2 Cr 2 O 7 solution for the oxidation to Na 2 SO 4 ⋅ Hence, molarity of Na 2 SO 3 solution is

Question

A 20.0 mL solution of Na 2 CO 3 required 30 mL of 0.01 M K 2 Cr 2 O 7 solution for the oxidation to Na 2 SO 4 ⋅ Hence, molarity of Na 2 SO 3 solution is (A) 0.015 M (B) 0.045 M (C) 0.030 M (D) 0.0225 M

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/chemistry/cccca5b20271b506791529bb7710d96d-.png" /> N1 V1( So 32-)=N2 V2( Cr 2 O 72-) 2 M1 × 20=6 × 0.01 × 30 M1=0.045 M

Q. A $20.0 m L$ solution of $N a_{2} C O_{3}$ required $30 m L$ of $0.01 M K_{2} C r_{2} O_{7}$ solution for the oxidation to $N a_{2} S O_{4} \cdot$ Hence, molarity of $N a_{2} S O_{3}$ solution is

$0.015 M$

$0.045 M$

$0.030 M$

$0.0225 M$

$N_{1} V_{1} (S o_{3}^{2 -}) = N_{2} V_{2} (C r_{2} O_{7}^{2 -})$
$2 M_{1} \times 20 = 6 \times 0.01 \times 30$
$M_{1} = 0.045 M$

Q. A 20.0mL solution of Na2​CO3​ required 30mL of 0.01MK2​Cr2​O7​ solution for the oxidation to Na2​SO4​⋅ Hence, molarity of Na2​SO3​ solution is

0.015M

0.045M

0.030M

0.0225M