HELP BEING TIMEDDDD 1.The solubility limit of glucose in water is 91g / 100mL of
ID: 821298 • Letter: H
Question
HELP BEING TIMEDDDD
1.The solubility limit of glucose in water is 91g / 100mL of water at 25°C. You try to dissolve 802g of glucose in 955 mL of water at 25°C. How much of your glucose will dissolve in your experiment
2.What is the concentration of LiCl (lithium chloride) when 2.5g of LiCl are dissolved in water and the volume of the final solution is 39.0mL? The formula for %(m/v) is shown below.
3What is the concentration of a liquid where 7.00 grams of NaCl are dissolved using 25.0 grams of water? Use units of %(m/m).
The formula for % (m/m) is
4,How many grams of K2CO3 are in 750mL of a 3.5% (m/v) solution of K2CO3? The formula for % (m/v) is shown below:
78.1% (m/m)
21.9% (m/m)
4.57% (m/m)
3.57% (m/m)
28.0% (m/m)
5.How many grams of K2CO3 are in 750mL of a 3.5% (m/v) solution of K2CO3? The formula for % (m/v) is shown below:
6.What is the mass (
78.1% (m/m)
21.9% (m/m)
4.57% (m/m)
3.57% (m/m)
28.0% (m/m)
Explanation / Answer
1) 91 grams of glucose is disssolved in -------------------------100 ml of water
802 grams of glucose is dissolved in ----------------------------ml of water
=802*100/91
=881.318 ml of water is needed.
but we are providing 955 ml of water.hence total glucose is didsolved in water.
2)( m/v)% = mass of the LiCl*100/volume of the solution
=2.5*100/39
=6.410%
3)(m/m)%= weight of the NaCl*100/total mass of the solution
=7*100/25
=28%
4)
( m/v)% = mass of the K2CO3l*100/volume of the solution
3.5%=mass of theOF THE K2CO3*100/750
MASS OF THE K2CO3=26.25%
5)
( m/v)% = mass of the K2CO3l*100/volume of the solution
3.5%=mass of theOF THE K2CO3*100/750
MASS OF THE K2CO3=26.25%
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