SS2: CHEMISTRY - 2ND TERM
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Water & Solution I | Week 110 Topics|1 Quiz
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Water, Solution and Solubility | Week 2 & 39 Topics|1 Quiz
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Air | Week 44 Topics|1 Quiz
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Pollution | Week 56 Topics|1 Quiz
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Rate of Chemical Reaction | Week 6 & 76 Topics|1 Quiz
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Energy and Energy Effect I | Week 8 & 97 Topics|1 Quiz
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Energy and Energy Effect II | Week 10 & 116 Topics|1 Quiz
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Chemical Equilibrium | Week 128 Topics|1 Quiz
Calculations on Solubility
Topic Content:
- Calculations on Solubility
Example 2.5.1:
30 cm3 of a saturated solution of a salt NaX was dissolved in water at 25°C, was evaporated to dryness and only 2.5 g of solid was left. Calculate the solubility of the salt in water at 25°C in moles/dm3 (Na = 23, X = 19)
Solution
30 cm3 of saturated solution contains 2.5 g
1000 cm3 of the solution will contain
\( \frac{1000\:cm^3}{30\:cm^3} \: \times \: \frac{2.5g}{1} \\ \normalsize = \scriptsize 83.3\: g/dm^3 \)Solubility in mol/dm3 = \( \frac{Solubility\: in \: g/dm^3}{Molar\:Mass} \)
Molar mass of NaX = 23 + 19 = 42 g/mol
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Thank you so much for these questions and solutions. It helped me practice my ability to solve questions on solubility.
Thanks very much…but help us check the site…for the calculation
Thank you so much ❤️💓💓
Thank you for the explanation. Very useful
Thank you for the calculation
I understand solubility now🙏🙏🙏🥰😍
Thanks slot, the calculation make me understood more on solubility