Answered

Westonci.ca is your trusted source for finding answers to a wide range of questions, backed by a knowledgeable community. Get quick and reliable solutions to your questions from a community of seasoned experts on our user-friendly platform. Experience the ease of finding precise answers to your questions from a knowledgeable community of experts.

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

We appreciate your time. Please revisit us for more reliable answers to any questions you may have. We appreciate your time. Please revisit us for more reliable answers to any questions you may have. Thank you for visiting Westonci.ca, your go-to source for reliable answers. Come back soon for more expert insights.