306866
Normality of 0.25 M phosphorous acid \(\mathrm{H}_{3} \mathrm{PO}_{3}\) is
1 0.125 N
2 0.75 N
3 0.50 N
4 0.25 N
Explanation:
We know that, Normality \(=\) Molarity \(\times\) Valency Since phosphorous acid \(\left(\mathrm{H}_{3} \mathrm{PO}_{3}\right)\) is dibasic thus, has valency 2 . \(\therefore\) Normality \(=0.25 \mathrm{M} \times 2=0.50 \mathrm{~N}\)
CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY
306867
1.25g of a solid dibasic acid is completely neutralised by 25 mL of 0.25 molar \({\rm{Ba(OH}}{{\rm{)}}_{\rm{2}}}\) solution. Molecular mass of the acid is
306868
\(\mathrm{\mathrm{x}}\) moles of potassium dichromate oxidises 1 mole of ferrous oxalate in acidic medium. Here \(\mathrm{\mathrm{x}}\) is
1 3
2 1.5
3 0.5
4 1.0
Explanation:
The reaction of oxidation of ferrous oxalate by potassium dichromate in acidic medium is written as \({\rm{2Fe}}{{\rm{C}}_{\rm{2}}}{{\rm{O}}_{\rm{4}}}{\rm{ + C}}{{\rm{r}}_{\rm{2}}}{\rm{O}}_{\rm{7}}^{{\rm{2 - }}}{\rm{ + 14}}{{\rm{H}}^{\rm{ + }}} \to \) \({\rm{2F}}{{\rm{e}}^{{\rm{3 + }}}}{\rm{ + 2C}}{{\rm{r}}^{{\rm{3 + }}}}{\rm{ + 4C}}{{\rm{O}}_{\rm{2}}}{\rm{ + 7}}{{\rm{H}}_{\rm{2}}}{\rm{O}}\) \(\mathrm{\because 2}\) moles of \(\mathrm{\mathrm{FeC}_{2} \mathrm{O}_{4}}\) are oxidised by \(\mathrm{=1}\) mole of \(\mathrm{\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}}\). \(\mathrm{\therefore 1}\) mole of \(\mathrm{\mathrm{FeC}_{2} \mathrm{O}_{4}}\) will be oxidised by \(\mathrm{=\dfrac{1}{2}=0.5}\) mole of \(\mathrm{\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}}\)
AIIMS - 2010
CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY
306869
One gram of hydrogen is found to combine with 80 g of bromine. One gram of calcium (valency = 2) combines with 4 g of bromine. The equivalent weight of calcium is
1 10
2 20
3 40
4 80
Explanation:
As 1 g \({{\rm{H}}_{\rm{2}}}\) combines with 80 g \({\rm{B}}{{\rm{r}}_{\rm{2}}}\), eq. wt. of \({\rm{Br = 80}}\). 4 g \({\rm{B}}{{\rm{r}}_{\rm{2}}}\) combine with 1 g Ca \(\therefore {\mkern 1mu} {\mkern 1mu} {\rm{80}}{\mkern 1mu} {\mkern 1mu} {\rm{g}}{\mkern 1mu} {\mkern 1mu} {\rm{B}}{{\rm{r}}_{\rm{2}}}\) combine with Ca = 20 g Hence eq. wt. of Ca = 20.
306866
Normality of 0.25 M phosphorous acid \(\mathrm{H}_{3} \mathrm{PO}_{3}\) is
1 0.125 N
2 0.75 N
3 0.50 N
4 0.25 N
Explanation:
We know that, Normality \(=\) Molarity \(\times\) Valency Since phosphorous acid \(\left(\mathrm{H}_{3} \mathrm{PO}_{3}\right)\) is dibasic thus, has valency 2 . \(\therefore\) Normality \(=0.25 \mathrm{M} \times 2=0.50 \mathrm{~N}\)
CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY
306867
1.25g of a solid dibasic acid is completely neutralised by 25 mL of 0.25 molar \({\rm{Ba(OH}}{{\rm{)}}_{\rm{2}}}\) solution. Molecular mass of the acid is
306868
\(\mathrm{\mathrm{x}}\) moles of potassium dichromate oxidises 1 mole of ferrous oxalate in acidic medium. Here \(\mathrm{\mathrm{x}}\) is
1 3
2 1.5
3 0.5
4 1.0
Explanation:
The reaction of oxidation of ferrous oxalate by potassium dichromate in acidic medium is written as \({\rm{2Fe}}{{\rm{C}}_{\rm{2}}}{{\rm{O}}_{\rm{4}}}{\rm{ + C}}{{\rm{r}}_{\rm{2}}}{\rm{O}}_{\rm{7}}^{{\rm{2 - }}}{\rm{ + 14}}{{\rm{H}}^{\rm{ + }}} \to \) \({\rm{2F}}{{\rm{e}}^{{\rm{3 + }}}}{\rm{ + 2C}}{{\rm{r}}^{{\rm{3 + }}}}{\rm{ + 4C}}{{\rm{O}}_{\rm{2}}}{\rm{ + 7}}{{\rm{H}}_{\rm{2}}}{\rm{O}}\) \(\mathrm{\because 2}\) moles of \(\mathrm{\mathrm{FeC}_{2} \mathrm{O}_{4}}\) are oxidised by \(\mathrm{=1}\) mole of \(\mathrm{\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}}\). \(\mathrm{\therefore 1}\) mole of \(\mathrm{\mathrm{FeC}_{2} \mathrm{O}_{4}}\) will be oxidised by \(\mathrm{=\dfrac{1}{2}=0.5}\) mole of \(\mathrm{\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}}\)
AIIMS - 2010
CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY
306869
One gram of hydrogen is found to combine with 80 g of bromine. One gram of calcium (valency = 2) combines with 4 g of bromine. The equivalent weight of calcium is
1 10
2 20
3 40
4 80
Explanation:
As 1 g \({{\rm{H}}_{\rm{2}}}\) combines with 80 g \({\rm{B}}{{\rm{r}}_{\rm{2}}}\), eq. wt. of \({\rm{Br = 80}}\). 4 g \({\rm{B}}{{\rm{r}}_{\rm{2}}}\) combine with 1 g Ca \(\therefore {\mkern 1mu} {\mkern 1mu} {\rm{80}}{\mkern 1mu} {\mkern 1mu} {\rm{g}}{\mkern 1mu} {\mkern 1mu} {\rm{B}}{{\rm{r}}_{\rm{2}}}\) combine with Ca = 20 g Hence eq. wt. of Ca = 20.
306866
Normality of 0.25 M phosphorous acid \(\mathrm{H}_{3} \mathrm{PO}_{3}\) is
1 0.125 N
2 0.75 N
3 0.50 N
4 0.25 N
Explanation:
We know that, Normality \(=\) Molarity \(\times\) Valency Since phosphorous acid \(\left(\mathrm{H}_{3} \mathrm{PO}_{3}\right)\) is dibasic thus, has valency 2 . \(\therefore\) Normality \(=0.25 \mathrm{M} \times 2=0.50 \mathrm{~N}\)
CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY
306867
1.25g of a solid dibasic acid is completely neutralised by 25 mL of 0.25 molar \({\rm{Ba(OH}}{{\rm{)}}_{\rm{2}}}\) solution. Molecular mass of the acid is
306868
\(\mathrm{\mathrm{x}}\) moles of potassium dichromate oxidises 1 mole of ferrous oxalate in acidic medium. Here \(\mathrm{\mathrm{x}}\) is
1 3
2 1.5
3 0.5
4 1.0
Explanation:
The reaction of oxidation of ferrous oxalate by potassium dichromate in acidic medium is written as \({\rm{2Fe}}{{\rm{C}}_{\rm{2}}}{{\rm{O}}_{\rm{4}}}{\rm{ + C}}{{\rm{r}}_{\rm{2}}}{\rm{O}}_{\rm{7}}^{{\rm{2 - }}}{\rm{ + 14}}{{\rm{H}}^{\rm{ + }}} \to \) \({\rm{2F}}{{\rm{e}}^{{\rm{3 + }}}}{\rm{ + 2C}}{{\rm{r}}^{{\rm{3 + }}}}{\rm{ + 4C}}{{\rm{O}}_{\rm{2}}}{\rm{ + 7}}{{\rm{H}}_{\rm{2}}}{\rm{O}}\) \(\mathrm{\because 2}\) moles of \(\mathrm{\mathrm{FeC}_{2} \mathrm{O}_{4}}\) are oxidised by \(\mathrm{=1}\) mole of \(\mathrm{\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}}\). \(\mathrm{\therefore 1}\) mole of \(\mathrm{\mathrm{FeC}_{2} \mathrm{O}_{4}}\) will be oxidised by \(\mathrm{=\dfrac{1}{2}=0.5}\) mole of \(\mathrm{\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}}\)
AIIMS - 2010
CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY
306869
One gram of hydrogen is found to combine with 80 g of bromine. One gram of calcium (valency = 2) combines with 4 g of bromine. The equivalent weight of calcium is
1 10
2 20
3 40
4 80
Explanation:
As 1 g \({{\rm{H}}_{\rm{2}}}\) combines with 80 g \({\rm{B}}{{\rm{r}}_{\rm{2}}}\), eq. wt. of \({\rm{Br = 80}}\). 4 g \({\rm{B}}{{\rm{r}}_{\rm{2}}}\) combine with 1 g Ca \(\therefore {\mkern 1mu} {\mkern 1mu} {\rm{80}}{\mkern 1mu} {\mkern 1mu} {\rm{g}}{\mkern 1mu} {\mkern 1mu} {\rm{B}}{{\rm{r}}_{\rm{2}}}\) combine with Ca = 20 g Hence eq. wt. of Ca = 20.
306866
Normality of 0.25 M phosphorous acid \(\mathrm{H}_{3} \mathrm{PO}_{3}\) is
1 0.125 N
2 0.75 N
3 0.50 N
4 0.25 N
Explanation:
We know that, Normality \(=\) Molarity \(\times\) Valency Since phosphorous acid \(\left(\mathrm{H}_{3} \mathrm{PO}_{3}\right)\) is dibasic thus, has valency 2 . \(\therefore\) Normality \(=0.25 \mathrm{M} \times 2=0.50 \mathrm{~N}\)
CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY
306867
1.25g of a solid dibasic acid is completely neutralised by 25 mL of 0.25 molar \({\rm{Ba(OH}}{{\rm{)}}_{\rm{2}}}\) solution. Molecular mass of the acid is
306868
\(\mathrm{\mathrm{x}}\) moles of potassium dichromate oxidises 1 mole of ferrous oxalate in acidic medium. Here \(\mathrm{\mathrm{x}}\) is
1 3
2 1.5
3 0.5
4 1.0
Explanation:
The reaction of oxidation of ferrous oxalate by potassium dichromate in acidic medium is written as \({\rm{2Fe}}{{\rm{C}}_{\rm{2}}}{{\rm{O}}_{\rm{4}}}{\rm{ + C}}{{\rm{r}}_{\rm{2}}}{\rm{O}}_{\rm{7}}^{{\rm{2 - }}}{\rm{ + 14}}{{\rm{H}}^{\rm{ + }}} \to \) \({\rm{2F}}{{\rm{e}}^{{\rm{3 + }}}}{\rm{ + 2C}}{{\rm{r}}^{{\rm{3 + }}}}{\rm{ + 4C}}{{\rm{O}}_{\rm{2}}}{\rm{ + 7}}{{\rm{H}}_{\rm{2}}}{\rm{O}}\) \(\mathrm{\because 2}\) moles of \(\mathrm{\mathrm{FeC}_{2} \mathrm{O}_{4}}\) are oxidised by \(\mathrm{=1}\) mole of \(\mathrm{\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}}\). \(\mathrm{\therefore 1}\) mole of \(\mathrm{\mathrm{FeC}_{2} \mathrm{O}_{4}}\) will be oxidised by \(\mathrm{=\dfrac{1}{2}=0.5}\) mole of \(\mathrm{\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}}\)
AIIMS - 2010
CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY
306869
One gram of hydrogen is found to combine with 80 g of bromine. One gram of calcium (valency = 2) combines with 4 g of bromine. The equivalent weight of calcium is
1 10
2 20
3 40
4 80
Explanation:
As 1 g \({{\rm{H}}_{\rm{2}}}\) combines with 80 g \({\rm{B}}{{\rm{r}}_{\rm{2}}}\), eq. wt. of \({\rm{Br = 80}}\). 4 g \({\rm{B}}{{\rm{r}}_{\rm{2}}}\) combine with 1 g Ca \(\therefore {\mkern 1mu} {\mkern 1mu} {\rm{80}}{\mkern 1mu} {\mkern 1mu} {\rm{g}}{\mkern 1mu} {\mkern 1mu} {\rm{B}}{{\rm{r}}_{\rm{2}}}\) combine with Ca = 20 g Hence eq. wt. of Ca = 20.
