5.3 % (W/V) Na 2 CO 3 solution and 6.3 % (W/V) H 2 C 2 O 4 ⋅ 2 H 2 O solution have same

Question

5.3 % (W/V) Na 2 CO 3 solution and 6.3 % (W/V) H 2 C 2 O 4 ⋅ 2 H 2 O solution have same (A) molarity (B) molality (C) normality (D) mole fraction

Tardigrade Admin · Accepted Answer

M Na 2 CO 3=( w B × 1000/ m B × V ) M H 2 C 2 O 4 .2 H 2 O =( w B × 1000/ m B × V ) =(5.3 × 1000/106 × 100)=0.5 =(6.3 × 1000/126 × 1000)=0.5 N Na 2 CO 3=( w B × 1000/ E B × V ) N H 2 C 2 O 4 .2 H 2 O =( W B × 1000/ E B × V ) =(5.3 × 1000/53 × 100)=1 =(6.3 × 1000/63 × 100)=1

Q. 5.3% (W/V) Na2​CO3​ solution and 6.3% (W/V) H2​C2​O4​⋅2H2​O solution have same

molarity

molality

normality

mole fraction

MNa2​CO3​​=mB​×VwB​×1000​ MH2​C2​O4​.2H2​O​=mB​×VwB​×1000​ =106×1005.3×1000​=0.5 =126×10006.3×1000​=0.5 NNa2​CO3​​=EB​×VwB​×1000​ NH2​C2​O4​.2H2​O​=EB​×VWB​×1000​ =53×1005.3×1000​=1 =63×1006.3×1000​=1

Q. $5.3%$ (W/V) $N a_{2} C O_{3}$ solution and $6.3%$ (W/V) $H_{2} C_{2} O_{4} \cdot 2 H_{2} O$ solution have same