163840
If \(\alpha\) is the degree of dissociation of \(\mathrm{Na}_2 \mathrm{SO}_4\) the van't Hoff's factor (i) used for calculating the molecular mass is :
163841
What will molality of 1 liter solution of \(93 \% \mathrm{H}_2 \mathrm{SO}_4\) (weight/volume) The density of the solution \(1.84 \mathrm{~g}\) Iml:
1 \(10.43 \mathrm{~m}\)
2 \(0.1043 \mathrm{~m}\)
3 \(104.3 \mathrm{~m}\)
4 \(0.00143 \mathrm{~m}\)
Explanation:
Mass of \(\mathrm{H}_2 \mathrm{SO}_4\) in \(100 \mathrm{ml}\) of \(93 \% \mathrm{H}_2 \mathrm{SO}_4\) solution = \(93 \mathrm{~g}\) mass of \(\mathrm{H}_2 \mathrm{SO}_4\) in \(1000 \mathrm{ml}\) of the \(\mathrm{H}_2 \mathrm{SO}_4\) solution \(930 \mathrm{~g}\) mass of \(1000 \mathrm{ml} \mathrm{H}_2 \mathrm{SO}_4\) solution \(=1000 \times 1.84\) \(=1840 \mathrm{~g}\) Mass of water in \(1000 \mathrm{ml}\) solution \(1840-930=910 \mathrm{~g}=0.910 \mathrm{~kg}\) Mole of \(\mathrm{H}_2 \mathrm{SO}_4=\frac{\text { wt. of } \mathrm{H}_2 \mathrm{SO}_4}{\text { Mol. wt. of } \mathrm{H}_2 \mathrm{SO}_4}=\frac{930}{98}\) Moles of \(\mathrm{H}_2 \mathrm{SO}_4\) in \(1 \mathrm{Kg}\) of water \( =\frac{930}{98} \times \frac{1}{0.910}=10.43 \mathrm{M} . \)
NCERT-XII-03
4 RBTS PAPER
163842
Which will form maximum boiling azeotrope :
\(\mathrm{HNO}_3+\mathrm{H}_2 \mathrm{O}\) solution shows negative deviation from ideal behaviour. Hence, it forms azeotrope with maximum boiling point.
163840
If \(\alpha\) is the degree of dissociation of \(\mathrm{Na}_2 \mathrm{SO}_4\) the van't Hoff's factor (i) used for calculating the molecular mass is :
163841
What will molality of 1 liter solution of \(93 \% \mathrm{H}_2 \mathrm{SO}_4\) (weight/volume) The density of the solution \(1.84 \mathrm{~g}\) Iml:
1 \(10.43 \mathrm{~m}\)
2 \(0.1043 \mathrm{~m}\)
3 \(104.3 \mathrm{~m}\)
4 \(0.00143 \mathrm{~m}\)
Explanation:
Mass of \(\mathrm{H}_2 \mathrm{SO}_4\) in \(100 \mathrm{ml}\) of \(93 \% \mathrm{H}_2 \mathrm{SO}_4\) solution = \(93 \mathrm{~g}\) mass of \(\mathrm{H}_2 \mathrm{SO}_4\) in \(1000 \mathrm{ml}\) of the \(\mathrm{H}_2 \mathrm{SO}_4\) solution \(930 \mathrm{~g}\) mass of \(1000 \mathrm{ml} \mathrm{H}_2 \mathrm{SO}_4\) solution \(=1000 \times 1.84\) \(=1840 \mathrm{~g}\) Mass of water in \(1000 \mathrm{ml}\) solution \(1840-930=910 \mathrm{~g}=0.910 \mathrm{~kg}\) Mole of \(\mathrm{H}_2 \mathrm{SO}_4=\frac{\text { wt. of } \mathrm{H}_2 \mathrm{SO}_4}{\text { Mol. wt. of } \mathrm{H}_2 \mathrm{SO}_4}=\frac{930}{98}\) Moles of \(\mathrm{H}_2 \mathrm{SO}_4\) in \(1 \mathrm{Kg}\) of water \( =\frac{930}{98} \times \frac{1}{0.910}=10.43 \mathrm{M} . \)
NCERT-XII-03
4 RBTS PAPER
163842
Which will form maximum boiling azeotrope :
\(\mathrm{HNO}_3+\mathrm{H}_2 \mathrm{O}\) solution shows negative deviation from ideal behaviour. Hence, it forms azeotrope with maximum boiling point.
163840
If \(\alpha\) is the degree of dissociation of \(\mathrm{Na}_2 \mathrm{SO}_4\) the van't Hoff's factor (i) used for calculating the molecular mass is :
163841
What will molality of 1 liter solution of \(93 \% \mathrm{H}_2 \mathrm{SO}_4\) (weight/volume) The density of the solution \(1.84 \mathrm{~g}\) Iml:
1 \(10.43 \mathrm{~m}\)
2 \(0.1043 \mathrm{~m}\)
3 \(104.3 \mathrm{~m}\)
4 \(0.00143 \mathrm{~m}\)
Explanation:
Mass of \(\mathrm{H}_2 \mathrm{SO}_4\) in \(100 \mathrm{ml}\) of \(93 \% \mathrm{H}_2 \mathrm{SO}_4\) solution = \(93 \mathrm{~g}\) mass of \(\mathrm{H}_2 \mathrm{SO}_4\) in \(1000 \mathrm{ml}\) of the \(\mathrm{H}_2 \mathrm{SO}_4\) solution \(930 \mathrm{~g}\) mass of \(1000 \mathrm{ml} \mathrm{H}_2 \mathrm{SO}_4\) solution \(=1000 \times 1.84\) \(=1840 \mathrm{~g}\) Mass of water in \(1000 \mathrm{ml}\) solution \(1840-930=910 \mathrm{~g}=0.910 \mathrm{~kg}\) Mole of \(\mathrm{H}_2 \mathrm{SO}_4=\frac{\text { wt. of } \mathrm{H}_2 \mathrm{SO}_4}{\text { Mol. wt. of } \mathrm{H}_2 \mathrm{SO}_4}=\frac{930}{98}\) Moles of \(\mathrm{H}_2 \mathrm{SO}_4\) in \(1 \mathrm{Kg}\) of water \( =\frac{930}{98} \times \frac{1}{0.910}=10.43 \mathrm{M} . \)
NCERT-XII-03
4 RBTS PAPER
163842
Which will form maximum boiling azeotrope :
\(\mathrm{HNO}_3+\mathrm{H}_2 \mathrm{O}\) solution shows negative deviation from ideal behaviour. Hence, it forms azeotrope with maximum boiling point.
163840
If \(\alpha\) is the degree of dissociation of \(\mathrm{Na}_2 \mathrm{SO}_4\) the van't Hoff's factor (i) used for calculating the molecular mass is :
163841
What will molality of 1 liter solution of \(93 \% \mathrm{H}_2 \mathrm{SO}_4\) (weight/volume) The density of the solution \(1.84 \mathrm{~g}\) Iml:
1 \(10.43 \mathrm{~m}\)
2 \(0.1043 \mathrm{~m}\)
3 \(104.3 \mathrm{~m}\)
4 \(0.00143 \mathrm{~m}\)
Explanation:
Mass of \(\mathrm{H}_2 \mathrm{SO}_4\) in \(100 \mathrm{ml}\) of \(93 \% \mathrm{H}_2 \mathrm{SO}_4\) solution = \(93 \mathrm{~g}\) mass of \(\mathrm{H}_2 \mathrm{SO}_4\) in \(1000 \mathrm{ml}\) of the \(\mathrm{H}_2 \mathrm{SO}_4\) solution \(930 \mathrm{~g}\) mass of \(1000 \mathrm{ml} \mathrm{H}_2 \mathrm{SO}_4\) solution \(=1000 \times 1.84\) \(=1840 \mathrm{~g}\) Mass of water in \(1000 \mathrm{ml}\) solution \(1840-930=910 \mathrm{~g}=0.910 \mathrm{~kg}\) Mole of \(\mathrm{H}_2 \mathrm{SO}_4=\frac{\text { wt. of } \mathrm{H}_2 \mathrm{SO}_4}{\text { Mol. wt. of } \mathrm{H}_2 \mathrm{SO}_4}=\frac{930}{98}\) Moles of \(\mathrm{H}_2 \mathrm{SO}_4\) in \(1 \mathrm{Kg}\) of water \( =\frac{930}{98} \times \frac{1}{0.910}=10.43 \mathrm{M} . \)
NCERT-XII-03
4 RBTS PAPER
163842
Which will form maximum boiling azeotrope :
\(\mathrm{HNO}_3+\mathrm{H}_2 \mathrm{O}\) solution shows negative deviation from ideal behaviour. Hence, it forms azeotrope with maximum boiling point.
