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goblinko [34]
3 years ago
13

How many grams of Nacl would you need to prepare 200 ml of a 5 M SOLUTION. ?

Chemistry
1 answer:
vazorg [7]3 years ago
8 0

Answer:

58.44 g of NaCl are needed.

Explanation:

Given data:

Mass of NaCl needed = ?

Volume of solution = 200 mL (200/1000 =0.2 L)

Molarity of solution = 5 M

Solution:

We will solve this problem through molarity formula.

Molarity is used to describe the concentration of solution. It tells how many moles are dissolve in per litter of solution.

Formula:

Molarity = number of moles of solute / L of solution

Now we will put the values.

5 M = moles of solute / 0.2 L

Moles of solute = 5 mol/L × 0.2 L

Moles of solute = 1 mol

Mass of sodium chloride:

Mass = number of moles × molar mass

Mass = 1 mol × 58.44 g/mol

Mass = 58.44 g

Thus, 58.44 g of NaCl needed.

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Why melting point of Fe in blast furnace is lower than ellingham?
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How many grams are in 3.14 moles of PI₃?
OverLord2011 [107]

Answer:

\boxed {\boxed {\sf 1290 \ g \ PI_3}}

Explanation:

We want to convert from moles to grams, so we must use the molar mass.

<h3>1. Molar Mass</h3>

The molar mass is the mass of 1 mole of a substance. It is the same as the atomic masses on the Periodic Table, but the units are grams per mole (g/mol) instead of atomic mass units (amu).

We are given the compound PI₃ or phosphorus triiodide. Look up the molar masses of the individual elements.

  • Phosphorus (P): 30.973762 g/mol
  • Iodine (I): 126.9045 g/mol

Note that there is a subscript of 3 after the I in the formula. This means there are 3 moles of iodine in 1 mole of the compound PI₃. We should multiply iodine's molar mass by 3, then add phosphorus's molar mass.

  • I₃: 126.9045 * 3=380.7135 g/mol
  • PI₃: 30.973762 + 380.7135 = 411.687262 g/mol

<h3>2. Convert Moles to Grams</h3>

Use the molar mass as a ratio.

\frac {411.687262 \ g \ PI_3}{ 1 \  mol \ PI_3}

We want to convert 3.14 moles to grams, so we multiply by that value.

3.14 \ mol \ PI_3 *\frac {411.687262 \ g \ PI_3}{ 1 \  mol \ PI_3}

The units of moles of PI₃ cancel.

3.14 *\frac {411.687262 \ g \ PI_3}{ 1 }

1292.698 \ g\ PI_3

<h3>3. Round</h3>

The original measurement of moles has 3 significant figures, so our answer must have the same. For the number we calculated, that is the tens place.

  • 1292.698

The 2 in the ones place tells us to leave the 9.

1290 \ g \ PI_3

3.14 moles of phosphorous triiodide is approximately equal to <u>1290 grams of phosphorus triodide.</u>

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Explanation:

b

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