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galben [10]
3 years ago
5

A 49.48-mL sample of an ammonia solution is analyzed by titration with HCl. It took 38.73 mL of 0.0952 M HCl to titrate the ammo

nia. What is the concentration of the original ammonia solution?
Chemistry
1 answer:
Kryger [21]3 years ago
8 0

Answer:

M₂ = 0.0745 M

Explanation:

In case of titration , the following formula can be used -

M₁V₁ = M₂V₂

where ,

M₁ = concentration of acid ,

V₁ = volume of acid ,

M₂ = concentration of base,

V₂ = volume of base .

from , the question ,

M₁ = 0.0952 M

V₁ = 38.73 mL

M₂ = ?

V₂ = 49.48 mL

Using the above formula , the molarity of ammonia , can be calculated as ,

M₁V₁ = M₂V₂  

0.0952 M * 38.73 mL = M₂* 49.48 mL

M₂ = 0.0745 M

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What is the mass of 4.99×1021 platinum atoms?
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Answer:

\boxed {\boxed {\sf 1.62 \ g \ Pt}}

Explanation:

We are asked to find the mass of a number of platinum (Pt) atoms.

<h3>1. Convert Atoms to Moles </h3>

First, we convert atoms to moles using Avogadro's Number or 6.022 × 10²³. This is the number of particles (atoms, molecules, formula units, etc.) in 1 mole of a substance. In this case, the particles are platinum atoms.

We will convert using dimensional analysis so we set up a ratio using the information we know (6.022 × 10²³ platinum atoms in 1 mole of platinum).

\frac {6.022 \times 10^{23} \ atoms \ Pt}{ 1 \ mol \ Pt}

We are converting 4.99 ×10²¹ atoms of Pt to moles of Pt, so we multiply by this value.

4.99 \times 10^{21} \ atoms \ Pt *\frac {6.022 \times 10^{23} \ atoms \ Pt}{ 1 \ mol \ Pt}

Flip the ratio so the units of atoms of platinum cancel.

4.99 \times 10^{21} \ atoms \ Pt *\frac { 1 \ mol \ Pt}{6.022 \times 10^{23} \ atoms \ Pt}

4.99 \times 10^{21} *\frac { 1 \ mol \ Pt}{6.022 \times 10^{23}}

\frac { 4.99 \times 10^{21}}{6.022 \times 10^{23}} \ mol \ Pt

Divide.

0.008286283627 \ mol \ Pt

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

Next, we convert moles to grams using the molar mass. This is the mass of 1 mole of a substance. This is found on the Periodic Table because it is equivalent to the atomic mass, but the units are grams per mole instead of atomic mass units. Look up platinum's molar mass.

  • Pt: 195.08 g/mol

Set up another ratio using this new information (195.08 grams of Pt in 1 mole of Pt).

\frac {195.08 \ g \ Pt}{ 1 \ mol \ Pt}

Multiply by the number of moles we just calculated.

0.008286283627 \ mol \ Pt*\frac {195.08 \ g \ Pt}{ 1 \ mol \ Pt}

The units of moles of platinum cancel.

0.008286283627*\frac {195.08 \ g \ Pt}{ 1 }

0.008286283627* {195.08 \ g \ Pt}

1.61648821\ g \ Pt

<h3>3. Round</h3>

The original measurement of atoms ( 4.99 ×10²¹ ) has 3 significant figures, so our answer must have the same. For the number we found, that is the hundredth place. The 6 in the thousandth place tells us to round the 1 up to a 2.

1.62 \ g \ Pt

There are approximately <u>1.62 grams of platinum</u> in 4.99 ×10²¹ atoms of platinum.

8 0
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