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28 January, 02:22

A student used 10 mL water instead of 30 mL for extraction of salt from mixture. How may this change the percentage of NaCl extracted?

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  1. 28 January, 04:50
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    It will be extracted only 1/3 of NaCl less in 10 mL of water than in 30 mL of water.

    If it is known that solubility of NaCl is 360 g/L, let's find out how many NaCl is in 30 mL of water:

    360 g : 1 L = x g : 30 mL

    Since 1 L = 1,000 mL, then:

    360 g : 1,000 mL = x g : 30 mL

    Now, crossing the products:

    x · 1,000 mL = 360 g · 30 mL

    x · 1,000 mL = 10,800 g mL

    x = 10,800 g : 1,000

    x = 10.8 g

    So, from 30 mL mixture, 10.8 g of NaCl could be extracted.

    Let's calculate the same for 10 mL water instead of 30 mL.

    360 g : 1 L = x g : 10 mL

    Since 1 L = 1,000 mL, then:

    360 g : 1,000 mL = x g : 10 mL

    Now, crossing the products:

    x · 1,000 mL = 360 g · 10 mL

    x · 1,000 mL = 3,600 g mL

    x = 3,600 g : 1,000

    x = 3.6 g

    So, from 10 mL mixture, 3.6 g of NaCl could be extracted.

    Now, let's compare:

    If from 30 mL mixture, 10.8 g of NaCl could be extracted and from 10 mL mixture, 3.6 g of NaCl could be extracted, the ratio is:

    3.6/10.8 = 1/3

    Therefore, i t will be extracted only 1/3 of NaCl less in 10 mL of water than in 30 mL of water.
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