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24 May, 20:24

How would you prepare 175.0 no of. 150 M AgNO3 solution starting with pure AgNO3?

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  1. 24 May, 20:36
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    The weight of solute required = Molarity x Molecular mass x Volume of solution in mL / 1000

    =0.15 x 170 x 175 / 1000 = 4.4625g

    Take 4.4625g of AgNO3 and disolve in certain quantity of water (<175 mL) and finally make up the solution to 175mL with water.
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