The relative density of a mixture of nitrogen and oxygen is 14.4 (H = 1) and the relative densities of nitrogen and oxygen are 14.0 and 16.0 (H = 1) respectively. Calculate the composition of the mixture (i) by volume and (ii) by mass.
Just tell me what does "H=1" mean in this question!
Answers
Answer:
Explanation:
i) Let 1 mL of the mixture contain x mL of N
2
and (1 - x) mL of O
2
.
Mass of x mL of N
2
=140×x=14xg (Mass =d×V)
Mass of (1 - x) mL of O
2
=16.0×(1−x)=(16.0−16x)g
Total mass of mixture =14x+16.0−16x
So, 14x+16.0−16x=14.4×1=14.4
or x = 0.8 , i.e., 80% by volume
Oxygen =1−x=(1−0.8)=0.2, i.e., 20% by volume.
(ii) Let 1 g of the mixture contain x g of N
2
and (1- x) g of oxygen.
Volume of x g of N
2
=
14.0
x
(V=
density
mass
)
Volume of (1-x) g of O
2
=
16.0
(1−x)
Total volume of the mixture =
14.0
1
+
16.0
(1−x)
So, =
14.0
1
+
16.0
(1−x)
=
14.4
1
or x=0.7778, i.e., 77.78% by mass
Oxygen =(1−x)=(1−0.7778)=0.2222
i.e., 22.22% by mass.
mark me as brainliest.
Explanation:
Answer:
\begin{gathered}(i) \: Let \: 1 \: of \: the \: mixture \: contain \: x \: ml \: of \: N2 \: and \: (1 - x) mL of N2 \: = 140 \times x \: = 14xg \: (Mass = d \times v) \: \\ Mass of (1 - x) mL of \: O2 \: = 16.0 \times (1 - x) \: = (16.0 - 16x)g \\ Total mass of mixture = 14x + 16.0 - 16x = 14.4 \times 1 = 14.4 \\ (ii) Let 1 g of the mixture contain x g of and (1- x) g of N2 \: oxygen. < /p > < p > Volume of x g of \\ \: N2 \: = \frac{x}{14.0} \\ Volume of (1-x) g of \: O2 = \: \frac{(1 - x)}{16.0} \\ Total volume of the mixture = \: \frac{1}{14.0} + \: \frac{(1 - x)}{16.0} \\ So, \frac{1}{14.0} + \frac{(1 \times x)}{16.0} = \frac{1}{14.4} \\ or x=0.7778 \: example \: 77.78percent \: by \: mass \\ oxygen \\ = (1 - x) = (1 - 0.7778) = 0.2222 \: or \: 22.22percent \: by \: mass\end{gathered}
(i)Let1ofthemixturecontainxmlofN2and(1−x)mLof N2=140×x=14xg(Mass=d×v)
Massof(1−x)mLofO2=16.0×(1−x)=(16.0−16x)g
Totalmassofmixture=14x+16.0−16x=14.4×1=14.4
(ii)Let1gofthemixturecontainxgof and(1−x)gofN2oxygen. </p><p>Volumeofxgof
N2=
14.0
x
Volumeof(1−x)gofO2=
16.0
(1−x)
Totalvolumeofthemixture =
14.0
1
+
16.0
(1−x)
So,
14.0
1
+
16.0
(1×x)
=
14.4
1
or x=0.7778example77.78percentbymass
oxygen
=(1−x)=(1−0.7778)=0.2222or22.22percentbymass
Hope that's helpful :)