41.  

Some statements are given below based on the pictures. Identify true and false statements.

(i) ‘P’ and ‘Q’ both indicates precision and accuracy.

(ii)‘Q’ indicates precision and accuracy white ‘R’ indicates neither precision nor accuracy.

(iii)‘P’ indicates precision but not accuracy.

(iv)‘Q’ indicates both precision and accuracy

A. FTTT

B. TTTT

C. TTFT

D. FTFT

Answer : Option A

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42.  The normality of 0.3M phosphorous acid is (IIT JEE 1999)

A. 0.1

B. 0.9

C. 0.3

D. 0.6

Answer : Option D

Explanation :

phosphorous acid is a dibasic acid. Its structure is as follows :   

→ Normality= basicity x Molarity = 2 x 0.3 = 0.6   

       

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43.  An aqueous solution of 6.3g oxalic acid dihydrate is made upto 250 mL. The volume of 0.1 N NaOH required to completely neutralize 10 mL of this solution is

A. 40 mL

B. 20 mL

C. 10 mL

D. 4 mL

Answer : Option A

Explanation :

Equivalents of \(H_{2}C_{2}O_{4}.2H_{2}O\)  in 10ml = Equivalents of NaOH

\(\therefore \frac{6.3\times 10000}{63\times 250\times 0.1}=V\)

                                    = 40mL

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44.  The pair of the compounds in which both the metals are in the highest possible oxidation state is

A. \(\left [ Fe(CN)_{6} \right ]^{3-}\), \(\left [ Co(CN)_{6} \right ]^{3-}\)

B. \(CrO_{2}Cl_{2}\), \(MnO_{4}^{-}\)

C. \(TiO_{3}\), \(MnO_{2}\)

D. \(\left [ Co(CN)_{6} \right ]^{3-}\), \(MnO_{3}\)

Answer : Option B

Explanation :

The oxidation states of various metals are :

→ Fe = +3, Co+3

→ Cr = +6, Mn+7

→ Ti = +6, Mn+4

→ Co = +3, Mn+6

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45.  In the analysis of 0.0500 g sample of feldspar, a mixture of the chiorides of sodium and potassium is obtained, which weighs 0.1180 g. Subsequent treatment of the mixed chlorides with silver nitrate gives 0.2451g of silver chloride. What is the percentage of a sodium oxide and potassium oxide in feldspar?

A. 10.62% \(Na_{2}O\), 3.58% \(K_{2}O\)

B. 3.58% \(Na_{2}O\), 10.62% \(K_{2}O\)

C. 10.62% \(Na_{2}O\), 35.8% \(K_{2}O\)

D. 35.8% \(Na_{2}O\), 10.62% \(K_{2}O\)

Answer : Option B

Explanation :

Suppose amount of NaCl in the mixture = ‘x’ g

\(\therefore \) The amount of KCl in the mixture = (0.118 - x) g

\(NaCl+AgNO_{3}\rightarrow AgCl+NaNO_{3}\)

   58.5                        143.5

\(\therefore \) x                      \(\frac{143.5\times x}{58.5}g\) ...............(i)

Similarly AgCl obtained from KCl = \(\frac{143.5\times (0.118-x)}{74.5}g\) ...............(ii)

But (i) + (ii) = 0.2451 g (Given)

Amount of NaCl = 0.0338 g

Amount of KCl = 0.0842 g

Now, 2NaCl = \(Na_{2}O\)

           117         62

        0.0338       \(\frac{0.0338\times 62}{117}\) = 0.0179 g

\(\therefore \) % of \(Na_{2}O\) = \(\frac{0.0179\times 100}{0.5}\) = 3.58%

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46.  5.5 g of a mixture of \(FeSO_{4}.7H_{2}O\) and \(Fe_{2}(SO_{4})_{3}.9H_{2}O\) requires 5.4 mL of 0.1N \(KMnO_{4}\) solution for complete oxidation. Calculate the number of mole of \(Fe_{2}(SO_{4})_{3}.9H_{2}O\) in the mixture.

A. 0.0095

B. 0.15

C. 0.0952

D. 1.52

Answer : Option A

Explanation :

→ Weight of \(FeSO_{4}.7H_{2}O\) = \(\frac{5.4\times 0.1\times 278}{1000}\) = \(0.150g\)

→ Moles of \(Fe_{2}(SO_{4})_{3}.9H_{2}O\)  = \(\frac{5.35}{562}\) = \(0.0095\)

View Answer

47.  A compound contains 28% of nitrogen and 72% of a metal by weight. Three atoms of the metal combine with two atoms of nitrogen. Find the equivalent weight of the metal.

A. 12

B. 24

C. 36

D. 48

Answer : Option A

Explanation :

→ \(Equivalent\ weight=\frac{Atomic\ Weight}{Valency}\)

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48.  The density of a 3M \(Na_{2}S_{2}O_{3}\) solution is 1.25 g per mL, What is the molalities of \(Na^{+}\) and \(S_{2}O_{3}^{2-}\) ions ?

A. 3.865, 7.732

B. 7.732, 3.865

C. 1.933, 7.732

D. 7.732, 1.933

Answer : Option B

Explanation :

→ \(m=\frac{1000 M}{1000d-MM_{5}}\), Where M = Molarity of solution, d = density of solution, \(M_{5}\) = Molar mass of solute

\(\therefore m=\frac{1000\times 3}{1000\times 1.25-3\times 158}\)

= 3.865

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49.  Hemoglobin present in blood contain 3.72% by mass iron. Calculate the number of iron atoms in 2.0g of hemoglobin .

A. \(4.53\times 10^{26}\)

B. \(4.53\times 10^{23}\)

C. \(5.95\times 10^{19}\)

D. \(8\times 10^{20}\)

Answer : Option C

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50.  How many moles of magnesium phosphate, \(Mg_{3}(PO_{4})_{2}\) will contain 0.25 mole of oxygen atoms (AIEEE 2006)

A. 0.02

B. \(3.125\times 10^{-2}\)

C. \(1.25\times 10^{-2}\)

D. \(2.5\times 10^{-2}\)

Answer : Option B

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