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Dennis_Churaev [7]
2 years ago
5

A forensic chemist is given a white powder for analysis. She dissolves 0.50 g of the substance in 8.0 g of benzene. The solution

freezes at 3.9°C. Can the chemist conclude that the compound is cocaine (C17H21N04)? What assumptions are made in the analysis? The freezing point of benzene is 5.5°C.
Chemistry
1 answer:
Sholpan [36]2 years ago
3 0

Answer:

The given compound cannot be cocaine.

Explanation:

The chemist can comment on the nature of compound being cocaine or not from the depression in freezing point.

Depression in freezing point of is related to molality as:

Depression in freezing point = Kf X molality

Where

Kf = cryoscopic constant = 4.90°C/m

depression in freezing point = normal freezing point - freezing point of solution

depression in freezing point = 5.5-3.9 = 1.6°C

1.6°C = 4.90 X molality

molality=\frac{1.6}{4.90} = 0.327 m

we know that:

molality=\frac{moles}{mass of solvent(kg)}=\frac{moles}{0.008kg}

therefore

moles = 0.327X0.008 = 0.00261 mol

moles=\frac{mass}{molarmass}

molarmass=\frac{mass}{moles}=\frac{0.50}{0.00261}= 191.57g/mol

The molar mass of cocaine is 303.353

So the given compound cannot be cocaine.

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Convert 2.0 M of Phenobarbital sodium (MW: 254 g/mole) solution in water into % w/v and ratio strengths.
Travka [436]

Answer:

The concentration is 50,8 % w/v and radio strengths = 1,96.

Explanation:

Phenobarbital sodium is a medication that could treat insomnia, for example.

2,0 M of Phenobarbital sodium means 2 moles in 1L.

The concentration units in this case are %w/v that means 1g in 100 mL and ratio strengths that means  1g in <em>r</em> mL. Thus, 2 moles must be converted in grams with molar weight -254 g/mole- and liters to mililiters -1 L are 1000mL-. So:

2 moles × \frac{254 g}{1 mole}= 508 g of Phenobarbital sodium.

1 L ×\frac{1000 mL}{ 1 L} = 1000 mL of solution

Thus, % w/v is:

\frac{508 g}{1000 mL} × 100 = 50,8 % w/v

And radio strengths:

\frac{1000 mL}{508 g}  = 1,96. Thus, you have 1 g in 1,96 mL

I hope it helps!

5 0
2 years ago
Please help!!
Sati [7]

In a chemical reaction, the limiting reagent is the chemical being used up while the excess reactant is the chemical left after the reaction process.

Before calculating the limiting and excess reactant, it is important to balance the equation first by stoichiometry.

C25N3H30Cl + NaOH = C25N3H30OH + NaCl

Since the reaction is already balanced, we can now identify which is the limiting and excess reagent.

First, we need to determine the number of moles of each chemical in the equation. This is crucial for determining the limiting and excess reagent.

<span>Assuming that there is the same amount  of solution X for each reactant</span>

1.0 M NaOH ( X ) = 1.0 moles NaOH

1.00 x 10-5 M C25N3H30Cl ( X ) = 1.00 x 10-5 moles C25N3H30Cl

<span>The result showed that the crystal violet has lesser amount than NaOH. Thus, the limiting reactant in this chemical reaction is crystal violet and the excess reactant is NaOH.</span>

3 0
2 years ago
What is the freezing point of a 1.40 m aqueous solution of a nonvolatile un-ionized solute? (the freezing point depression const
Alina [70]
Depression is freezing point is a colligative property. It is mathematically expressed as ΔTf = Kf X m

where Kf = <span>freezing point depression constant = 1.86°c kg /mol (for water)
m = molality of solution = 1.40 m

</span>∴ ΔTf = Kf X m  = 1.86 X 1.40 = 2.604 oC

Now, for water freezing point = 0 oC

∴Freezing point of solution = -2.604 oC
6 0
2 years ago
If a chemist analyzes a 3.84g sample containing sand and table sugar, and recovers 1.43g of      sand, what  percent by mass of
disa [49]
3.84 - 1.43 = 2.41
2.41g of table sugar

% mass = ( (mass of element) / (total mass) ) * 100
% mass = (2.41 / 3.84) * 100
% mass = (0.6276) * 100
% mass = 62.76

62.76%
8 0
2 years ago
Read 2 more answers
Three solutions contain a certain acid. The first contains 10% acid, the second 30%, and the third 50%. A chemist wishes to use
Fittoniya [83]

Answer:

To prepare 50L of 32% solution you need: 11L of 30% solution, 22L of 50% solution and 17L of 10% solution.

Explanation:

A 32% solution of acid means 32L of acid per 100L of solution. As the chemist wants to make a solution using twice as much of the 50% solution as of the 30% solution it is possible to write:

2x*50% + x*30% + y*10% = 50L*32%

<em>130x + 10y = 1600 </em><em>(1)</em>

<em>-Where x are volume of 30% solution, 2x volume of 50% solution and y volume of 10% solution-</em>

Also, it is possible to write a formula using the total volume (50L), thus:

<em>2x + x +y = 50L</em>

<em>3x + y = 50L </em><em>(2)</em>

If you replace (2) in (1):

130x + 10(50-3x) = 1600

100x + 500 = 1600

100x = 1100

<em>x = 11L -Volume of 30% solution-</em>

2x = 22L -Volume of 50% solution-

50L - 22L - 11L = 17 L -Volume of 10% solution-

I hope it helps!

8 0
2 years ago
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