# Question-1

**Classify the following as pure substances or mixtures separate to pure substances into elements and compounds and divide the mixtures into homogeneous and heterogeneous categories:**

(i) Bronze (ii) Smoke (iii) Pencil lead (iv) Antenna rod

(i) Bronze (ii) Smoke (iii) Pencil lead (iv) Antenna rod

**Solution:**

(i) Bronze | = a mixture of Cu & Sn - Homogenous mixture. |

(ii) Dust | = a mixture of carbon particle and air - Heterogenous mixture. |

(iii) Pencil lead | = Pure element - Graphite. |

(iv) Antenna rod | = Aluminium. |

# Question-2

**Calculate number of moles for the following ?**

(i) 360 gms of H

(i) 360 gms of H

_{2}O (ii) 5.6 gms of Nitrogen (iii) 4 gms of NaOH.**Solution:**

(i) No. of moles of H

_{2}O = = = 20;

(ii) Number of moles of N_{2} = = 0.2;

(iii) Number of moles of NaOH = = 0.1.

# Question-3

**Calculate the number of moles of H**

_{2}SO_{4}present in 50 ml of 0.2m H_{2}SO_{4}.**Solution:**

0.2m means it contains 0.2 moles per litre.

1000 ml contains 0.2 moles at H_{2}SO_{4 }

âˆ´50 ml contains = x 50 = = 10 Ã—10^{-2} = 0.01 moles.

# Question-4

**What is molarity of a solution contain 5.84 gms of NaCl in 200 ml of solution?**

**Solution:**

Number of moles of NaCl = = = 0.1

Molarity =

Volume of solution in litre = = 0.2

Molarity of the solution = = = 0.5m.

# Question-5

**What is weight of NaOH present in 250 cc of a 2M solution?**

**Solution:**

2M of NaOH solution means it contains 2 moles of NaOH per litre

1000ml = 2 moles of NaOH

Number of moles of NaOH present in 250cc = = = 0.5

Mole = ; wt = mole x m.wt

âˆ´ Weight of NaOH present in 250 ml of solution = 0.5 x 40 = 20 gm.

/ Through Formula /

Weight of substance in 1 lit solution = Molarity x m.wt

Weight of NaOH present in 1 lit solution = 2 x 40 gms

âˆ´ Weight of NaOH present in 250cc solution = = 20 gms.

# Question-6

**In a reaction vessel 0.980 gm of H**

_{2}SO_{4}is required to be added for completing the reaction. How many millilitre of 0.05 M H_{2}SO_{4}solution should be added for this requirement?**Solution:**

0.05 M H

_{2}SO

_{4}means 0.05 moles of H

_{2}SO

_{4}pressure in 1000 ml;

1 litre of 0.05 m H_{2}SO_{4 }contains = molarity x m.wt = 0.05 x 98 = 5 x 10^{-2} x98 = 4.9 gms of H_{2}SO_{4}

_{âˆ´}1 ml contains = = 4.9 x10^{-3} gms of H_{2}SO_{4 }

4.9 x of gms of H_{2}SO_{4}^{ }= 1 ml of solution

_{âˆ´}0.980 gms of H_{2}SO_{4} = = = 2 x 10^{2} = 200 ml.

Through formula :

Wt. of H_{2}SO_{4}^{ }present in 1 lt = Molarity x m.wt

Wt.of H_{2}SO_{4} present in 1 lt of 0.05M = 0.05 x 98 gm = 98 x 5 x 10^{-2 }gms = 4.9 gms.

_{âˆ´}4.9 gms is present in 1000 ml

Hence, 0.980 gms is present in

=

= 200 ml.

# Question-7

**How much AgCl will be formed by adding 200ml of 5M HCl to the solution containing 1.7 gms of Ag No**

_{3}?**Solution:**

AgNO3 + HCl â†’ AgCl â†“ + HNO3

200 ml of 5M HCl contains 1.7g of AgNO3

âˆ´1000 ml of 1M HCl will contain = 1.7g

Hence 1M solution of HCl contains 1.7 g of AgNO3.

# Question-8

**Calculate the weight of HCl in 10 ml of con. HCl of density 1.2 gm L**

^{-1}container 35% HCl by weight. What is the molarity of the solution?**Solution:**

35% HCl means 35 gm of HCl are present in 100 gms of HCl solution.

Volume of 100 gms of given HCl solution = = = 83.3 ml

83.3 ml of con. HCl contains 35 gms of HCl

âˆ´10 ml of con. HCl contains = = 4.20 gms of HCl.

Molarity of the solution = (in litre)

=

= = 1.15 M.

# Question-9

**The molarity of con. HCl is 1.15M; what volume of con. HCl is required to make 1.00 of 0.1M HCl.**

**Solution:**

Known Unknown

V_{1} M_{1} = V_{2} M_{2 }

1000 ml x 0.1M = V_{2} x 1.15M

V_{2} = ml = = = 86.9 ml.

# Question-10

**Aluminium and Sulphuric acid react according to the reaction :**

If 0.5 mol Al are added to H

If 0.5 mol Al are added to H

_{2}SO_{4}solution containing 0.2 mole H_{2}SO_{4}, how many moles of H_{2}are produced.**Solution:**

As per equation, 2 moles Al reacts with 3 moles of H

_{2}SO

_{4}to produce 3 mole of H

_{2}.

0.5 ml Al reacts with 3 mole of H_{2}SO_{4 }moles of H_{2}.

0.5 ml of Al reacts with 0.2 mole of H_{2}SO_{4} = mole of H_{2 }

= moles of H_{2 }

= = 0.4 moles of H_{2}.