Vapour density of a gas is 22. Its molecular mass will be:
Molecular mass = 2 x vapour density
= 2 x 22
Therefore, molecular mass of a gas will be 44.
What is the mass of 0.20 mole of C2H5OH (ethanol)?
Molecular Mass of Ethanol
= 2(12) + 5 (1) + 16 + 1
= 24 + 5 + 17
Moles (n) = Given Mass (m)/Molecular Mass (M)
Thus, to = nM
= (0.2) (46)
Consider the reaction between hydrogen and oxygen gases to form water. Which of the following is/are not conserved in the reaction?
2H2(g) + O2(g) → 2H2O(l)
In the equation both side of reaction moles of atoms is conserved ie, both reactant and product contain 4H and 2O but there is 3 (2+1) moles of reactant molecules turns into 2 moles of product molecules.
Mass of sucrose C12H22O11 produced by mixing 84 gm of carbon, 12 gm of hydrogen and 56 lit. O2 at 1 atm & 273 K according to given reaction, is C(s) + H2(g) + O2(g) → C12H22O11 (s)
C = 84/12 = 7 mole
H2 = 12 g = 6 mole
O2 = 56/22.4 = 5/2 mole
12C + 11H2 + 11/2 O2 → C12H22O11
L.R. = O2
11/2 mole O2 produce 1 mole sucrose 5/2 mole O2 will for 5/11 mole sucrose mass of sucrose = 5/11 × (mol. mass)
= 5/11 × 342
= 155.45 g
Suppose the elements X and Y combine to form two compounds XY2 and X3Y2. When 0.1 mole of XY2 weighs 10 g and 0.05 mole of X3Y2 weighs 9 g, the atomic weights of X and Y are
Let atomic weight of x = Mx
atomic weight of y = My
mole = weight /atomic weight
a/c to question,
mole of xy2 = 0.1
0.1 = 10g/( Mx +2My)
Mx + 2My = 100g -------(1)
for x3y2 ; mole of x3y2 = 0.05
0.05 = 9/( 3Mx + 2My )
3Mx + 2My = 9/0.05 = 9× 20 = 180 g ---(2)
solve eqns (1) and (2)
2Mx = 80
Mx = 40g/mol
and My = 30g/mole
Number of atoms present in 52 g of helium is:
(NA is Avogadro’s constant Gram atomic mass of He is 4 g)
Gram atomic mass of helium is 4 g so number of atoms present in 52 g of helium
52/4 × NA = 13 NA
The vapour density of a mixture of gas A (Molecular mass = 40) and gas B (Molecular mass = 80) is 25.Then mole % of gas B in the mixture would be
As vapour density of mixture is 25,
Molar mass of mixture is 2×25=50.
Now, assume mole % of B in mixture is x.
Atomic mass of Cl is 35.5 g. Calculate the mass of 4.50 moles of chlorine gas, Cl2.
Mass of 1 mole of Cl2 is 71 g. Hence mass of 4.50 moles is 71 X 4.50 = 319.5 g
Atomic mass of bromine is 80 g. The mass of four moles of molecular bromine (Br2) is:
Mass of one mole of Br2 is 160g. Hence mass of four moles of molecular bromine is 4 X 160 = 640 g.
Number of nitrogen atoms present in 1.4 g of N2
Given weight of N₂ gas = 1.4 g
Molar mass of N₂ gas = 28 g
So, mole = given mass/ molar mass
⇒ mole = 1.4/28 = 1/20 mole
Now, number of molecules = mole × avogadro number
⇒ number of molecules = 1/20 × 6.022 × 10²³
⇒ number of molecules = 3.011 × 10²²
Now, we are asked for number of atoms. In N₂, there are 2 atoms, so to obtain number of atoms we will multiply with 2 in number of molecules.
⇒ number of atoms = 2 × 3.011 × 10²²
⇒ number of atoms = 6.022 × 10²²
Volume of 17g of NH3 at N.T.P. will be
Molar mass of NH3 is 17g. So the volume of 17g of NH3 i.e 1 mole NH3 at N.T.P. will be 22.4L.
Which of the following has maximum number of moles?
1g hydrogen has the maximum number of moles because atomic mass of hydrogen is smallest among all.
Molar mass of F2 is 38 g. How many atoms are present in 0.147 mole of F2?
One molecule of F2 contains 2 atoms of fluorine.
1 mol of fluorine contains 2 X 6.023 X 1023 atoms
So, 0.147 moles will contain 0.147 X 2 X 6.023 X 1023 = 1.76 X 1023 atoms of flourine.
Mass of 6.022 x 1022 molecules of CO2 is about:
1 mole of a substance is equal to its molecular mass expressed in terms of grams.
Gram molecular mass of CO2 = 12 + (16 Χ 2) = 12 + 32 = 44 g
Mass of 6.023 X1022 molecules of CO2 is 44g/10 = 4.4g.