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choli [55]
2 years ago
8

Ima 5% glucose solution, how many grams of glucose would be present per 100mL

Chemistry
1 answer:
Sedaia [141]2 years ago
6 0

Percentage Weight-in-volume is defined as the <em><u>number of grams of a solute in a 100 ml (milliliters) solution.</u></em>

<u />

<u>Percentage Weight-in-volume</u> can tell us about the <em>degree of concentration of a given solution.</em>

<em><u /></em>

The solute can be <em>crystalline or non-crystalline in nature.</em>

<em></em>

The <u>number of grams of glucose</u> present in a <u>5% glucose solution</u> is 5 grams.

  • This question is based on a Percentage Weight-in-volume. The formula states that:

a% of a glucose solution =<u> a grams of glucose in a 100 mL solution</u>

Hence, 5% glucose solution = 5 grams of glucose / 100 mL solution

Therefore, the <u>number of grams of glucose</u> present in a <u>5% glucose solution</u> is 5 grams.

To learn more, visit the link below:

brainly.com/question/8482854

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Which of the following is the greatest dissolved solid in ocean water? Magnesium Chloride Plankton Sodium Chloride Calium
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The right answer is Sodium chloride.

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3 years ago
A sample of N2(g) was collected over water at 25 oC and 730 torr in a container with a volume of 340 mL. The vapor pressure of w
harina [27]

Answer:

0.36 g of N2.

Explanation:

The following data were obtained from the question:

Temperature (T) = 25 °C

Volume (V) = 340 mL

Measured pressure = 730 torr

Vapour pressure = 23.76 torr

Mass of N2 =..?

First, we shall determine the true pressure of N2. This can be obtained as follow:

Measured pressure = 730 torr

Vapor pressure = 23.76 torr

True pressure =..?

True pressure = measured pressure – vapor pressure

True pressure = 730 – 23.76

True pressure = 706.24 torr.

Converting 706.24 torr to atm, we have:

760 torr = 1 atm

Therefore,

706.24 torr = 706.24 / 760 = 0.929 atm

Next, we shall convert 340 mL to L. This is illustrated below:

1000 mL = 1 L

Therefore,

340 mL = 340/1000 = 0.34 L

Next, we shall convert 25 °C to Kelvin temperature. This is illustrated below:

Temperature (K) = Temperature (°C) + 273

T(K) = T (°C) + 273

T (°C) = 25 °C

T(K) = 25 °C + 273

T (K) = 298 K

Next, we shall determine the number of mole of N2. This can be obtained as follow:

Pressure (P) = 0.929 atm

Volume (V) = 0.34 L

Temperature (T) = 298 K

Gas constant (R) = 0.0821 atm.L/Kmol

Number of mole (n) =...?

PV = nRT

0.929 x 0.34 = n x 0.0821 x 298

Divide both side by 0.0821 x 298

n = (0.929 x 0.34 ) /(0.0821 x 298)

n = 0.0129 mole

Finally, we shall determine the mass of N2 as shown below:

Mole of N2 = 0.0129 mole

Molar mass of N2 = 2x14 = 28 g/mol

Mass of N2 =.?

Mole = mass /Molar mass

0.0129 = mass of N2/ 28

Cross multiply

Mass of N2 = 0.0129 x 28

Mass of N2 = 0.36 g

Therefore, 0.36 g of N2 was collected.

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