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28 June, 10:13

How much water should be mixed with 237 ml of ammonia, whose strength is 100%, in order to create a mixture that is diluted to a 75% strength.

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  1. 28 June, 10:57
    0
    Let the amount of water to be mixed = x ml

    new mixture = 237+x

    The new mixture is diluted to 75% strength i. e. 75/100

    75/100 = 3/4

    Therefore,

    237 / (237+x) = 3/4

    4 * 237 = 3 (237+x)

    948=711+3x

    3x=948-711

    3x=237

    x=237/3

    x=79

    The amount of water to be mixed with ammonia = 79ml
  2. 28 June, 13:28
    0
    Let the x ml of water be mixed with 237 ml of ammonia whose strength is 100%, in order to create a mixture of 75% strength.

    The equation required to solve this will be:

    237 / (237+x) = 75/100

    solving for x we get:

    237 / (237+x) = 3/4

    237*4=3 (237+x)

    948=711+3x

    3x+711=948

    3x=948-711

    3x=237

    x=237/3

    x=79 ml

    therefore the amount of water to be added will be 79 ml
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