A gas is found to have a formula [CO]x. If its vapour density is 70, the value of x is:
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Solution
As we know that,Molecular mass of [CO]x= 2 × V.D.
⇨(12 + 16)x = 2 ×70 ⇨ 28x = 140 ⇨x =5
If 11/2 moles of oxygen combine with Al to form Al2O3the weight of Al used in the reaction is (Al = 27)
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Solution
In the reaction
4NH3(g) + 5O2(g) ⟶ 4NO(g) + 6H2O(l), when 1 mole of ammonia and 1 mole of O2 are made to react to completion
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Solution
In order to prepare one litre normal solution of KMnO4, how many grams of KMnO4 are required if the solution is to be used in acid medium for oxidation?
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Solution
Eq. wt of KMnO4 in acid medium is 31.6 g. Hence this much amount must be dissolved in 1 litre to prepare normal solution.
An organic compound whose empirical and molecular formula are same, contains 20% carbon, 6.7% hydrogen,46.7%nitrogen and the rest oxygen. On heating it yields ammonia,leaving a solid residue. The solid residue gives a violet colour with dilute solution of alkaline copper sulphate. The organic compound is
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Solution
A compound contains 54.55 % carbon, 9.09% hydrogen ,36.36% oxygen. The empirical formula of this compound is :
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Solution
The mass of BaCO3 produced when excess CO2 is bubbled through a solution of 0.205 mol Ba(OH)2 is:
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Solution
The empirical formula of a compound is CH2. One mole of this compound has a mass of 42 grams. Its molecular formula is :
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Solution
Empirical formula of compound = CH2 Molecular mass of the compound = 42
&therefore n = 42/14 = 3
∴ Hence molecular formula = C3H6
30 g of magnesium and 30 g of oxygen are reacted, then the residual mixture contains
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Solution
The number of atoms in 4.25 g of NH3 is approximately
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Solution
Number of atoms = \(\frac{4.25 \times 6.023 \times 10^{4} \times 4}{17}\) = 6 × 1023
(One molecule of NH3 contains 4 atoms 1 N and 3H)