Answered

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What mass of ammonium nitrate
is needed to dissolve in 995 g of
water to make a 1.50 m solution?

What Mass Of Ammonium Nitrate Is Needed To Dissolve In 995 G Of Water To Make A 150 M Solution class=

Sagot :

elvokipkorir

Answer:

=> 119.4597 g

Explanation:

From our question, we have been provided with the molarity of the resultant solution.

We know that, molarity of a solution is contained in 1 L or 1000 cm³ of solution.

Hence we have 1.50 moles in 1000 cm³.

Also, we know that, 1 cm³ equivalent to 1 g for water.

Thus, from our question we have 995 cm³ of water.

Lets now solve our problem;

1.5 moles are contained in 1000 cm³

x moles are contained in 995 cm³

Cross multiply

[tex]x \: moles \: = \frac{(995 {cm}^{3} \times 1.5 \: moles )}{(1000 {cm}^{3} )} \\ = \frac{1492.5 \: moles}{1000} \\ = 1.4925 \: moles[/tex]

1.4925 moles of NH4NO3 are required to be dissolved in water.

To find the mass of solute (NH4NO3) we use the formula;

Mass = RFM (Relative Formula Mass) x Number of moles

[tex] = 80.04 \: \frac{g}{mol} \times1.4925 \: mol \\ = 119.4597 \: g [/tex]

Therefore the mass of NH4NO3 needed to be dissolved in 995 g of water is 119.4597 g

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