Question

The table shows the solubility of two substances in water at 20 °C. Solubility of Substances Mass (kg) Solubility of Sodium Chloride (g per 100 g of water) Solubility of Lead Nitrate (g per 100 g of water) 100 L K 200 36 54 Part 1: What will be the values of L and K for 100 kg of each substance? Part 2: Explain your answer for Part 1.

Answer 1

(a) At 313 K 62 g of potassium nitrate dissolved in 100 g of water. So to produce a saturated solution of potassium nitrate in 50 g of water we need

100

62

×50=31 g of potassium nitrate.

(b)Some soluble potassium chloride will separate out in the form of crystal at room temperature because the solubility of potassium chloride will decrease.

(c)

(i) Solubility of Potassium nitrate at 293 K is 32 g.

(ii) Solubility of sodium chloride at 293 K is 36 g.

(iii) Solubility of Potassium chloride at 293 K is 35 g.

(iv) Solubility of Ammonium chloride at 293 K is 37 g.

The solubility of Ammonium chloride is highest at this temperature.

(d)The solubility of salt increases with the increase in temperature.

Answer 2

"Its seems this is what you are looking for"

The table shows the solubility of two substances in water at 20 °C.

Mass          Solubility of Sodium chloride        Solubility of Lead Nitrate

(Kg)              (g per 100 g of water)                    (g per 100 g of water)    

100                          L                                                        54

200                        36                                                       K

Part 1: What will be the values of L and K for of each substance?

Part 2: Explain your answer for Part 1.

Answer:

The value of K and L for each substance are 108g and 18g respectively.

Explanation:

We have given the solubility of NaCl in 100g of water = 36g

Solubility of Lead Nitrate in 100g of water = 54g

Calculate the value of L:-

Solubility of 200Kg of sodium chloride = 36g

Solubility of 100kg of Sodium chloride $=\frac{36g}{200Kg} *100Kg=18g$

Calculate the value of K:-

Solubility of 100Kg of Lead Nitrate = 54g

Solubility of 200kg of Lead Nitrate $=\frac{54g}{100Kg} *200Kg=108g$

Difference in the solubility = solubility of K - solubility of L

                                            =  108g - 18g = 90g

Therefore, from the value K and L, we can say that Lead Nitrate is more soluble than sodium chloride in water.