9.8g of solid potassium chlorate (KCIO₃) were added to 40 cm³ of water and heated to dissolve all the solid. As he solution cooled, crystals of potassium chlorate started forming at 60°C. The solubility of potassium chlorate at 66°C is therefore?

24.5 mole per dm³

0.2 mole per dm³

10.0 mole per dm³

20.0 mole per dm³

2.0 mole per dm³

Correct Option

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4 years ago

n=m/M
n=C×V/1000
m/M=C×V/1000
M=39+35.5+16×3=122.5g
9.8/122.5=C×40/1000
C=9.8×1000/122.5×40
C=9800/4900=2mole/dm³

3 years ago

I got 2.. Why do they want to confuse some one with all this temperature

1 year ago

I don't understand 😔

1 year ago

Sol.= m/mm × 1000/vol.

m=9.8
mm of KClO3= 39+35.5+3(16)
= 39+35.5+48=122.5
v=40

sol.= 9.8/122.5 × 1000/40

= 9800/4900

= 2

1 year ago

How do you guys manage to read 3 months before the exam😭😭😂🤲🏻
I reprinted before I started reading 😂

3 years ago

4 over 10
and im writing today by 12

ceoofwahala

20th June, 2026

Chemistry

2 comments

ASSAAS

20th June, 2026

English Language

5 comments

infinitehoaxx

21st May, 2026

Computer

4 comments

9.8g of solid potassium chlorate (KCIO3) were added to 40 cm3 of water and heated to dissolve all the solid. As he solution cooled, crystals of potassium chlorate started forming at 60°C. The solubility of potassium chlorate at 66°C is therefore?