205470
एक विलयन जिसमें द्रव्यमान के अनुसार \(25\% {H_2}O\), \(25\% {C_2}{H_5}OH\) एवं \(50\% \,C{H_3}\,COOH\) उपस्थित है। इसमें \({H_2}O\)के मोल प्रभाज होंगे
1 \(0.25\)
2 \(2.5\)
3 \(0.502\)
4 \(5.03\)
Explanation:
Mole fraction of \(H _{2} O =\frac{\text { number of moles of } H _{2} O }{\text { total number of moles of all components }}\) Let the total mass of solution \(=100 g\) Mass of \(H _{2} O =25 g\) Mass of \(C _{2} H _{5} OH =25 g\) Mass of \(CH _{3} COOH =50 g\) Moles of \(H _{2} O =\frac{25}{18}=1.388\) \(\left(\because \text { Molar mass of } H _{2} O =18\right)\) Moles of \(C _{2} H _{5} OH =\frac{25}{46}=0.543\) \(\left(\because \text { Molar mass of } C _{2} H _{5} OH =46\right)\) Moles of \(CH _{3} COOH =\frac{50}{60}=0.833\) \(\left(\because \text { Molar mass of } CH _{3} COOH =60\right)\) Total no. of moles \(=1.388+0.543+0.833=2.764\) \(\therefore\) Mole fraction of \(H _{2} O =\frac{1.388}{2.764}=0.502\)
02. SOLUTIONS (HM)
205471
\({H_2}S{O_4}\) का \(5\) मोलर विलयन \(1\) लीटर से \( 10\) लीटर तक तनु किया गया तो विलयन की नॉर्मलता ...... \(N\) होगी
205470
एक विलयन जिसमें द्रव्यमान के अनुसार \(25\% {H_2}O\), \(25\% {C_2}{H_5}OH\) एवं \(50\% \,C{H_3}\,COOH\) उपस्थित है। इसमें \({H_2}O\)के मोल प्रभाज होंगे
1 \(0.25\)
2 \(2.5\)
3 \(0.502\)
4 \(5.03\)
Explanation:
Mole fraction of \(H _{2} O =\frac{\text { number of moles of } H _{2} O }{\text { total number of moles of all components }}\) Let the total mass of solution \(=100 g\) Mass of \(H _{2} O =25 g\) Mass of \(C _{2} H _{5} OH =25 g\) Mass of \(CH _{3} COOH =50 g\) Moles of \(H _{2} O =\frac{25}{18}=1.388\) \(\left(\because \text { Molar mass of } H _{2} O =18\right)\) Moles of \(C _{2} H _{5} OH =\frac{25}{46}=0.543\) \(\left(\because \text { Molar mass of } C _{2} H _{5} OH =46\right)\) Moles of \(CH _{3} COOH =\frac{50}{60}=0.833\) \(\left(\because \text { Molar mass of } CH _{3} COOH =60\right)\) Total no. of moles \(=1.388+0.543+0.833=2.764\) \(\therefore\) Mole fraction of \(H _{2} O =\frac{1.388}{2.764}=0.502\)
02. SOLUTIONS (HM)
205471
\({H_2}S{O_4}\) का \(5\) मोलर विलयन \(1\) लीटर से \( 10\) लीटर तक तनु किया गया तो विलयन की नॉर्मलता ...... \(N\) होगी
205470
एक विलयन जिसमें द्रव्यमान के अनुसार \(25\% {H_2}O\), \(25\% {C_2}{H_5}OH\) एवं \(50\% \,C{H_3}\,COOH\) उपस्थित है। इसमें \({H_2}O\)के मोल प्रभाज होंगे
1 \(0.25\)
2 \(2.5\)
3 \(0.502\)
4 \(5.03\)
Explanation:
Mole fraction of \(H _{2} O =\frac{\text { number of moles of } H _{2} O }{\text { total number of moles of all components }}\) Let the total mass of solution \(=100 g\) Mass of \(H _{2} O =25 g\) Mass of \(C _{2} H _{5} OH =25 g\) Mass of \(CH _{3} COOH =50 g\) Moles of \(H _{2} O =\frac{25}{18}=1.388\) \(\left(\because \text { Molar mass of } H _{2} O =18\right)\) Moles of \(C _{2} H _{5} OH =\frac{25}{46}=0.543\) \(\left(\because \text { Molar mass of } C _{2} H _{5} OH =46\right)\) Moles of \(CH _{3} COOH =\frac{50}{60}=0.833\) \(\left(\because \text { Molar mass of } CH _{3} COOH =60\right)\) Total no. of moles \(=1.388+0.543+0.833=2.764\) \(\therefore\) Mole fraction of \(H _{2} O =\frac{1.388}{2.764}=0.502\)
02. SOLUTIONS (HM)
205471
\({H_2}S{O_4}\) का \(5\) मोलर विलयन \(1\) लीटर से \( 10\) लीटर तक तनु किया गया तो विलयन की नॉर्मलता ...... \(N\) होगी
205470
एक विलयन जिसमें द्रव्यमान के अनुसार \(25\% {H_2}O\), \(25\% {C_2}{H_5}OH\) एवं \(50\% \,C{H_3}\,COOH\) उपस्थित है। इसमें \({H_2}O\)के मोल प्रभाज होंगे
1 \(0.25\)
2 \(2.5\)
3 \(0.502\)
4 \(5.03\)
Explanation:
Mole fraction of \(H _{2} O =\frac{\text { number of moles of } H _{2} O }{\text { total number of moles of all components }}\) Let the total mass of solution \(=100 g\) Mass of \(H _{2} O =25 g\) Mass of \(C _{2} H _{5} OH =25 g\) Mass of \(CH _{3} COOH =50 g\) Moles of \(H _{2} O =\frac{25}{18}=1.388\) \(\left(\because \text { Molar mass of } H _{2} O =18\right)\) Moles of \(C _{2} H _{5} OH =\frac{25}{46}=0.543\) \(\left(\because \text { Molar mass of } C _{2} H _{5} OH =46\right)\) Moles of \(CH _{3} COOH =\frac{50}{60}=0.833\) \(\left(\because \text { Molar mass of } CH _{3} COOH =60\right)\) Total no. of moles \(=1.388+0.543+0.833=2.764\) \(\therefore\) Mole fraction of \(H _{2} O =\frac{1.388}{2.764}=0.502\)
02. SOLUTIONS (HM)
205471
\({H_2}S{O_4}\) का \(5\) मोलर विलयन \(1\) लीटर से \( 10\) लीटर तक तनु किया गया तो विलयन की नॉर्मलता ...... \(N\) होगी
