NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY
306850
The vapour density of a chloride of an element is 39.5 . The \({\mathrm{E w}}\) of the elements is 3.82 . The atomic weight of the element is ____ .
1 7.64
2 3.95
3 1
4 10
Explanation:
\({\mathrm{M w}}\) of metal chloride \({\mathrm{\left(\mathrm{MCl}_{x}\right)=M+x \times 35.5}}\) \(\begin{aligned}& =2 \times \mathrm{VD}=2 \times 39.5 \\& =79.0\end{aligned}\) \({\mathrm{M w=E w \times x}}\) Valency of metal \({\mathrm{=x}}\) Atomic weight of element \({\mathrm{=E w \times x}}\) \({\mathrm{\mathrm{MCl}_{x}=3.82 \times x+x \times 35.5=79}}\) \({\mathrm{x=\dfrac{79}{3.82+35.5} \approx 2}}\) Atomic weight \({\mathrm{=E w \times x=3.82 \times 2=7.64}}\)
CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY
306851
Match Column I with Column II and choose the correct combination from the given options given.
(A) Molarity & (D) normality depend on temperature, dilution, and volume i.e.,A-P,Q,R;D-P,Q,R (B) Molality & (C) mole dfraction depend only on dilution i.e.,B-Q;C-Q. So, correct option is (1).
CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY
306852
When a metal is burnt, its weight is increased by 24 per cent. The equivalent weight of the metal will be
1 2
2 24
3 33.3
4 76
Explanation:
If \(100 \mathrm{~g}\) of metal is burnt, then it combines with oxygen to form metal oxide. As its weight increases by \(24 \%\), then the mass of combined oxygen would be 24 g. Thus, the equivalent weight of metal = \(\frac{{{\rm{weight}}\,{\rm{of}}\,{\rm{metal}}}}{{{\rm{weight}}\,{\rm{of}}\,{\rm{oxygen}}\,{\rm{combined}}}} \times 8\) \( = \frac{{100}}{{24}} \times 8\) \( = 33.33\;{\rm{g}}\,{\rm{eq}}\,{\rm{mo}}{{\rm{l}}^{{\rm{ - 1}}}}\) Hence, the equivalent weight of metal will be 33.3.
CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY
306853
The number of molecules in \(100 \mathrm{~mL}\) of \({\rm{0}}{\rm{.02}}\,{\rm{N}}{{\rm{H}}_{\rm{2}}}{\rm{S}}{{\rm{O}}_{\rm{4}}}\) is
306850
The vapour density of a chloride of an element is 39.5 . The \({\mathrm{E w}}\) of the elements is 3.82 . The atomic weight of the element is ____ .
1 7.64
2 3.95
3 1
4 10
Explanation:
\({\mathrm{M w}}\) of metal chloride \({\mathrm{\left(\mathrm{MCl}_{x}\right)=M+x \times 35.5}}\) \(\begin{aligned}& =2 \times \mathrm{VD}=2 \times 39.5 \\& =79.0\end{aligned}\) \({\mathrm{M w=E w \times x}}\) Valency of metal \({\mathrm{=x}}\) Atomic weight of element \({\mathrm{=E w \times x}}\) \({\mathrm{\mathrm{MCl}_{x}=3.82 \times x+x \times 35.5=79}}\) \({\mathrm{x=\dfrac{79}{3.82+35.5} \approx 2}}\) Atomic weight \({\mathrm{=E w \times x=3.82 \times 2=7.64}}\)
CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY
306851
Match Column I with Column II and choose the correct combination from the given options given.
(A) Molarity & (D) normality depend on temperature, dilution, and volume i.e.,A-P,Q,R;D-P,Q,R (B) Molality & (C) mole dfraction depend only on dilution i.e.,B-Q;C-Q. So, correct option is (1).
CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY
306852
When a metal is burnt, its weight is increased by 24 per cent. The equivalent weight of the metal will be
1 2
2 24
3 33.3
4 76
Explanation:
If \(100 \mathrm{~g}\) of metal is burnt, then it combines with oxygen to form metal oxide. As its weight increases by \(24 \%\), then the mass of combined oxygen would be 24 g. Thus, the equivalent weight of metal = \(\frac{{{\rm{weight}}\,{\rm{of}}\,{\rm{metal}}}}{{{\rm{weight}}\,{\rm{of}}\,{\rm{oxygen}}\,{\rm{combined}}}} \times 8\) \( = \frac{{100}}{{24}} \times 8\) \( = 33.33\;{\rm{g}}\,{\rm{eq}}\,{\rm{mo}}{{\rm{l}}^{{\rm{ - 1}}}}\) Hence, the equivalent weight of metal will be 33.3.
CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY
306853
The number of molecules in \(100 \mathrm{~mL}\) of \({\rm{0}}{\rm{.02}}\,{\rm{N}}{{\rm{H}}_{\rm{2}}}{\rm{S}}{{\rm{O}}_{\rm{4}}}\) is
306850
The vapour density of a chloride of an element is 39.5 . The \({\mathrm{E w}}\) of the elements is 3.82 . The atomic weight of the element is ____ .
1 7.64
2 3.95
3 1
4 10
Explanation:
\({\mathrm{M w}}\) of metal chloride \({\mathrm{\left(\mathrm{MCl}_{x}\right)=M+x \times 35.5}}\) \(\begin{aligned}& =2 \times \mathrm{VD}=2 \times 39.5 \\& =79.0\end{aligned}\) \({\mathrm{M w=E w \times x}}\) Valency of metal \({\mathrm{=x}}\) Atomic weight of element \({\mathrm{=E w \times x}}\) \({\mathrm{\mathrm{MCl}_{x}=3.82 \times x+x \times 35.5=79}}\) \({\mathrm{x=\dfrac{79}{3.82+35.5} \approx 2}}\) Atomic weight \({\mathrm{=E w \times x=3.82 \times 2=7.64}}\)
CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY
306851
Match Column I with Column II and choose the correct combination from the given options given.
(A) Molarity & (D) normality depend on temperature, dilution, and volume i.e.,A-P,Q,R;D-P,Q,R (B) Molality & (C) mole dfraction depend only on dilution i.e.,B-Q;C-Q. So, correct option is (1).
CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY
306852
When a metal is burnt, its weight is increased by 24 per cent. The equivalent weight of the metal will be
1 2
2 24
3 33.3
4 76
Explanation:
If \(100 \mathrm{~g}\) of metal is burnt, then it combines with oxygen to form metal oxide. As its weight increases by \(24 \%\), then the mass of combined oxygen would be 24 g. Thus, the equivalent weight of metal = \(\frac{{{\rm{weight}}\,{\rm{of}}\,{\rm{metal}}}}{{{\rm{weight}}\,{\rm{of}}\,{\rm{oxygen}}\,{\rm{combined}}}} \times 8\) \( = \frac{{100}}{{24}} \times 8\) \( = 33.33\;{\rm{g}}\,{\rm{eq}}\,{\rm{mo}}{{\rm{l}}^{{\rm{ - 1}}}}\) Hence, the equivalent weight of metal will be 33.3.
CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY
306853
The number of molecules in \(100 \mathrm{~mL}\) of \({\rm{0}}{\rm{.02}}\,{\rm{N}}{{\rm{H}}_{\rm{2}}}{\rm{S}}{{\rm{O}}_{\rm{4}}}\) is
306850
The vapour density of a chloride of an element is 39.5 . The \({\mathrm{E w}}\) of the elements is 3.82 . The atomic weight of the element is ____ .
1 7.64
2 3.95
3 1
4 10
Explanation:
\({\mathrm{M w}}\) of metal chloride \({\mathrm{\left(\mathrm{MCl}_{x}\right)=M+x \times 35.5}}\) \(\begin{aligned}& =2 \times \mathrm{VD}=2 \times 39.5 \\& =79.0\end{aligned}\) \({\mathrm{M w=E w \times x}}\) Valency of metal \({\mathrm{=x}}\) Atomic weight of element \({\mathrm{=E w \times x}}\) \({\mathrm{\mathrm{MCl}_{x}=3.82 \times x+x \times 35.5=79}}\) \({\mathrm{x=\dfrac{79}{3.82+35.5} \approx 2}}\) Atomic weight \({\mathrm{=E w \times x=3.82 \times 2=7.64}}\)
CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY
306851
Match Column I with Column II and choose the correct combination from the given options given.
(A) Molarity & (D) normality depend on temperature, dilution, and volume i.e.,A-P,Q,R;D-P,Q,R (B) Molality & (C) mole dfraction depend only on dilution i.e.,B-Q;C-Q. So, correct option is (1).
CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY
306852
When a metal is burnt, its weight is increased by 24 per cent. The equivalent weight of the metal will be
1 2
2 24
3 33.3
4 76
Explanation:
If \(100 \mathrm{~g}\) of metal is burnt, then it combines with oxygen to form metal oxide. As its weight increases by \(24 \%\), then the mass of combined oxygen would be 24 g. Thus, the equivalent weight of metal = \(\frac{{{\rm{weight}}\,{\rm{of}}\,{\rm{metal}}}}{{{\rm{weight}}\,{\rm{of}}\,{\rm{oxygen}}\,{\rm{combined}}}} \times 8\) \( = \frac{{100}}{{24}} \times 8\) \( = 33.33\;{\rm{g}}\,{\rm{eq}}\,{\rm{mo}}{{\rm{l}}^{{\rm{ - 1}}}}\) Hence, the equivalent weight of metal will be 33.3.
CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY
306853
The number of molecules in \(100 \mathrm{~mL}\) of \({\rm{0}}{\rm{.02}}\,{\rm{N}}{{\rm{H}}_{\rm{2}}}{\rm{S}}{{\rm{O}}_{\rm{4}}}\) is
