Chemistry, asked by dharapkshitij, 4 days ago

What mass of salt T is needed to make 10 g of saturated solution at 343K?

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Answered by pushpalatasingh760
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Answer:

(I) At 323 K, salt Y has the highest solubility in water while salt Z has the lowest solubility

(I) At 323 K, salt Y has the highest solubility in water while salt Z has the lowest solubility(ii) By definition of saturated solution,

(I) At 323 K, salt Y has the highest solubility in water while salt Z has the lowest solubility(ii) By definition of saturated solution,100 g of water at 323 K contain salt = 40 g

(I) At 323 K, salt Y has the highest solubility in water while salt Z has the lowest solubility(ii) By definition of saturated solution,100 g of water at 323 K contain salt = 40 g125 g of water at 323 K contain salt = (40g)(100g)×(125g)=50g

(I) At 323 K, salt Y has the highest solubility in water while salt Z has the lowest solubility(ii) By definition of saturated solution,100 g of water at 323 K contain salt = 40 g125 g of water at 323 K contain salt = (40g)(100g)×(125g)=50g∴ Mass of salt to be added to make the solution again saturated = (50 - 40) = 10 g

(I) At 323 K, salt Y has the highest solubility in water while salt Z has the lowest solubility(ii) By definition of saturated solution,100 g of water at 323 K contain salt = 40 g125 g of water at 323 K contain salt = (40g)(100g)×(125g)=50g∴ Mass of salt to be added to make the solution again saturated = (50 - 40) = 10 g(iii) The data shows that the solubility of the salt Y is least affected with increase in temperature.

(I) At 323 K, salt Y has the highest solubility in water while salt Z has the lowest solubility(ii) By definition of saturated solution,100 g of water at 323 K contain salt = 40 g125 g of water at 323 K contain salt = (40g)(100g)×(125g)=50g∴ Mass of salt to be added to make the solution again saturated = (50 - 40) = 10 g(iii) The data shows that the solubility of the salt Y is least affected with increase in temperature.(iv) At 290 K, mass of T required to make the saturated in 200 g of water = (25g)(100g)×(200g)50g

Explanation:

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