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Reptile [31]
3 years ago
14

What is the molarity of a solution that contains 40. grams of NaOH in 0.50 liter of solution

Chemistry
1 answer:
statuscvo [17]3 years ago
5 0

Answer:

The molarity of the solution is: 2 M

Explanation:

We calculate the weight of 1 mol of NaOH:

Weight 1 mol NaOH= Weight Na + Weight 0 + Weight H= 23 g+16g+ 1 g= 40g/mol

The molarity is the quantity of moles in 1000ml (1 liter) of solution:

0,5liter-----40 grams NaOH

1 liter----x=(1 liter x 40 grams NOH)/0,5 liter = 80 grams NaOH

40 grams----1 mol NaOH

80 grams----X=(80 grams x 1 mol NaOH)/40 grams= 2 mol NaOH--> 2M

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Match the following terms with their definitions:
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If a student mixes 75 mL of 1.30 M HNO3 and 150 mL of 6.5 M NaOH. is the final solution acidic, basic, or neutral
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Answer:

The solution is basic.

Explanation:

We can determine the nature of the solution via determining which has the large no. of millimoles (acid or base):

  • If no. of millimoles of acid > that of base; the solution is acidic.
  • If no. of millimoles of acid = that of base; the solution is neutral.
  • If no. of millimoles of acid < that of base; the solution is basic.

  • We need to calculate the no. of millimoles of acid and base:

no. of millimoles of acid (HNO₃) = MV = (1.3 M)(75.0 mL) = 97.5 mmol.

no. of millimoles of base (NaOH) = MV = (6.5 M)(150.0 mL) = 975.0 mmol.

<em>∴ The no. of millimoles of base (NaOH) is larger by 10 times than the acid (HNO₃).</em>

<em>So, the solution is: basic.</em>

4 0
3 years ago
If you place 1.0 L of ethanol (C2H5OH) in a small laboratory that is 3.0 m long, 2.0 m wide, and 2.0 m high, will all the alcoho
ankoles [38]

If you place 1.0 L of ethanol (C2H5OH) in a small laboratory that is 3.0 m long, 2.0 m wide, and 2.0 m high, will all the alcohol evaporate? If some liquid remains, how much will there be? The vapor pressure of ethyl alcohol at 25 °C is 59 mm Hg, and the density of the liquid at this temperature is 0.785g/cm^3 .

will all the alcohol evaporate? or none at all?

Answer:

Yes, all the ethanol present in the laboratory will evaporate since the mole of ethanol present in vapor is greater. The volume of ethanol left will therefore  be zero.

Explanation:

Given that:

The volume of alcohol which is placed in a small laboratory = 1.0 L

Vapor pressure of ethyl alcohol  at 25 ° C = 59 mmHg

Converting 59 mmHg to atm ; since 1 atm = 760 mmHg;

Then, we have:

= \frac{59}{760}atm

= 0.078 atm

Temperature = 25 ° C

= ( 25 + 273 K)

= 298 K.

Density of the ethanol = 0.785 g/cm³

The volume of laboratory = l × b × h

= 3.0 m × 2.0 m × 2.5 m

= 15 m³

Converting the volume of laboratory to liter;

since 1 m³ = 100 L; Then, we  have:

15 × 1000 = 15,000 L

Using ideal gas equation to determine the moles of ethanol in vapor phase; we have:

PV = nRT

Making n the subject of the formula; we have:

n = \frac{PV}{RT}

n = \frac{0.078 * 15000}{0.082*290}

n = 47. 88 mol of ethanol

Moles of ethanol in 1.0 L bottle can be calculated as follows:

Since  numbers of moles = \frac{mass}{molar mass}

and mass = density × vollume

Then; we can say ;

number of moles = \frac{density*volume }{molar mass of ethanol}

number of moles =\frac{0.785g/cm^3*1000cm^3}{46.07g/mol}

number of moles = \frac{&85}{46.07}

number of moles = 17.039 mol

Thus , all the ethanol present in the laboratory will evaporate since the mole of ethanol present in vapor is greater. The volume of ethanol left will therefore be zero.

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