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NEET
NEET Chemistry MCQ
Mcq On Some Basic Concepts Of Chemistry
Quiz 1
Solution
Q.1.
The value of $2.01 \times .011$ using significant figure is
15%
.023
38%
.022
43%
.02211
4%
.0222
Q.2.
The value of $\frac {1.0042 \times .0034}{1.23}$ using significant figure is
35%
.0028
19%
.003
22%
.002
24%
.00279
Q.3.
841.45 to four significant figures will be
7%
841.0
15%
841.40
33%
841.5
45%
841.4
Q.4.
Which is of these zero is not significant.
7%
.901
55%
.011
35%
.1240
4%
.309
Q.5.
The value of $5396 \times .045 + 325.3$ is
11%
565
53%
568.3
30%
568
6%
560
Q.6.
The value of 70.3 - 1.245 using significant figure is
48%
69.1
14%
69.06
26%
69.055
12%
69.05
Q.7.
The value of 10.5+1.51+2.401 using significant figure is
17%
14.5
12%
14.41
21%
14.411
50%
14.4
Q.8.
Match the column
9%
p -> i, q -> iii, r -> iv, s-> ii
6%
p -> i, q -> iii, r -> ii, s-> iv
61%
p -> ii, q -> iii, r -> ii, s-> i
23%
p -> ii, q -> iii, r -> i, s-> ii
Q.9.
The scientific notation for the number 0.00000060 is
25%
$6 \times 10^{-7}$
53%
$6.0 \times 10^{-7}$
3%
$6.00 \times 10^{-7}$
18%
$60 \times 10^{-8}$
Q.10.
The significant figures in the number 1200 is
62%
2
4%
5
4%
3
31%
4
Q.11.
70.770 to three significant figures will be
23%
70.7
5%
70.0
11%
71.0
61%
70.8
Q.12.
which of these is not having 3 significant figures
26%
11100
17%
4.01
31%
1.410
26%
1.00
Q.13.
12.1756 to 5 significant figures
78%
12.176
6%
12.180
7%
12.175
9%
12.170
Q.14.
The value of $1.23 \times .231$ is
12%
.280
21%
0.285
40%
0.284
27%
0.28413
Q.15.
The value of 415.5 + 3.64 + .238 is
16%
419.378
52%
419.4
28%
419.38
4%
420
Q.16.
A solution of glucose in water is labelled as 10% w/w, The molality is
16%
.74 moles/kg
27%
.5 moles/kg
38%
0.62 moles/kg
20%
none of these
Q.17.
Which of the following terms are unit less?
63%
Mole fraction
29%
Formality
6%
Molality
2%
Molarity
Q.18.
Molarity of Solution having 5 moles of solutes present in 2 L of solution is
78%
2.5 M
15%
4 M
2%
.5 M
4%
2 M
Q.19.
The correct expression relating molality(m) , Molarity(M), density (d) and Molar Mass(MB) of solute is
20%
$m = \frac { d - MM_B }{ M}$
35%
$m = \frac {M}{d - MM_B}$
24%
$m = \frac {M}{d + MM_B}$
22%
$m = \frac { d + MM_B }{M }$
Q.20.
The molality of the solution which contains 10 g of Cane sugar($C_{12}H_{22}O_{11}$) dissolved in 150 g of water?
17%
.22 moles/kg
29%
1.1 moles/kg
10%
.5 moles/kg
44%
.194 moles/kg
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