Grade-XI M.M.-20
Subject-Chemistry Time-1hr.
Exam-Pre-mid(July-2020)
1. Calculate number of significant figures in 0.0671×10-34 : (1)
(a) 3 (b) 7 (c) 4 (d) None of these.
2. Assertion-Reason: Use the following data key to answer correctly (1)
(a) If both assertion & reason are correct,reason is correct explanation of assertion.
(b) If both assertion & reason are correct,reason is not correct explanation of assertion .
(c) If assertion is correct but reason is incorrect.
(d) If both assertion and reason are incorrect.
Assertion: 24 carat gold is homogeneous mixture.
Reason: A compound has fixed composition of elements present in it.
3. Explain the term molarity with its units. (1)
4. Write down the drawbacks of law of conservation of mass. (1)
5. Define the term mole & give its mathematical form. (1)
6. 0.79 g of iron reacts with 0.4 g sulphur to produce iron sulphide. If 2.8 g of iron is mixed with dil. HCl
and H2S is passed through mixture 4.4 g of iron sulphide is precipitated. Show that this data is according
to law of constant Composition. (2)
7. Calculate the no. of moles present in 1.6 ×10-19 Molecules of CO2 , also find their mass. (2 )
8. Calculate the molarity and normality of solution Containing 0.4 g NaoH in 500 ml of water. (3)
9. The molecular mass of benzene is 78 amu. Determine its molecular formula if its percentage is
C = 92.31% & H= 7.69%.(C=12amu, H=1amu) (3)
10. Calcium Carbonate reacts with aqueous HCL to give CaCl2 and CO2 as
CaCO3 + HCl → CaCl2 + H20 + CO2 .
What mass of CaCO3 is required react completely with 25ml of 0.75 molar HCl. (5)
bhi help in chemistry paper
Answers
Ans1. 3
Ans2. a
Ans3. Molarity is defined as the ratio of no. Of moles of solute per unit volume of solution in litres.
M=n/V
n= no. Of moles of solute and its unit is mole
V= volume of solution and its unit is litre
So molarity's unit = mol/litre
Ans4. Law of conservation of mass is not applicable for nuclear reactions because in that case some mass is changes into energy
Ans5. The quantity/weight/volume of atoms/molecules can be find in term of moles. Mole is the unit of atoms
For quantity
1 mole = avagadro's number
n=q/N.
Q=Quantity of atom
N= avagadro number
n= no. Of moles
For weight
1 mole=atomic/molecular mass
n=w/M
w= weight of substance
M=atomic./mecular mass
n= no. Of moles
For volume
1 mole= 22.7 litre
n=V/22.7
V= volume of gas
n= no. Of moles
Ans6.yes it follow law of chemical composition
Ans7. n=1.6 ×10-19/6.23 ×10-23
n=0.2568x10⁴
n=w/M
M of co2=44
nxM=w
0.2568x10⁴x44
11.03x10⁴
Ans8.M=n/V
n=w/M
n=0.4/40 molar mass of naoh
n=0.01
M=0.01x1000/500
M=0.02
Ans9.