1)
<span>m(NaCl) = 1.95 g
V(H2O) = 250mL
M(NaCl) = </span><span>58.5 g/mole
Since waters density value is 1g/mL, it can be assumed that volume and mass of water are same values:
</span>V(H2O) = 250ml = 250g = 0.25 kg<span>
</span><span>molality of NaCl:
</span><span>
n(NaCl)=m/M=1.95/58.5= 0.033 mole
</span>molality b(NaCl)=n(NaCl) / V (H2O)= 0.033/0.25 = 0.132 mol/kg
<span>
milimolality of NaOH = 0.132/0,001 = 132 mmole/kg
</span>
milliosmolality of NaOH = milimolality x N of ions formed in dissociation
Since NaCl dissociates into 2 ions in solution:
<span>
</span>milliosmolality of NaOH = 132 x 2 = 264 osmol<span>es/kg
</span>
2)
m(gl) = 9 g
V(H2O) = 250mL
M(NaCl) = 180 g/mole
Since waters density value is 1g/mL, it can be assumed that volume and mass of water are same values:
V(H2O) = 250ml = 250g = 0.25 kg
molality of glucose:
n(gl)=m/M=9/180= 0.05 mole
molality b(gl)=n(gl) / V (H2O)= 0.05/0.25 = 0.2 mol/kg
milimolality of glucose = 0.132/0,001 = 200 mmole/kg
milliosmolality of glucose = milimolality x N of ions formed in dissociation
Since glucose does not dissociate, milimolality and milliosmolality are same:
milliosmolality of glucose = 200 osmoles/kg
3)
The osmosis represents the diffusion of solvent molecules through a semi-permeable membrane that allows passage solvent molecules but does not to the dissolved substance molecule. The osmosis occurs when the concentrations of the solution on both sides of the membrane are different. Since the semi-permeable membrane only permeates the solvent molecules, but not the particles of the dissolved substance, it occurs the solvent diffusion through the membrane, i.e. the solvent molecules pass through the membrane to equalize the concentration on both sides of the membrane. Solvents molecules move from the middle with a lower concentration in the middle with a higher concentration of dissolved substances.
In our case, osmosis will occur because the concentration of NaCl solution and the concentration of glucose solution do not have same values. Osmosis will occur in the direction of glucose solution because it has a lower concentration.
Answer:
The metal has a heat capacity of 0.385 J/g°C
This metal is copper.
Explanation:
<u>Step 1</u>: Data given
Mass of the metal = 21 grams
Volume of water = 100 mL
⇒ mass of water = density * volume = 1g/mL * 100 mL = 100 grams
Initial temperature of metal = 122.5 °C
Initial temperature of water = 17°C
Final temperature of water and the metal = 19 °C
Heat capacity of water = 4.184 J/g°C
<u />
<u>Step 2: </u>Calculate the specific heat capacity
Heat lost by the metal = heat won by water
Qmetal = -Qwater
Q = m*c*ΔT
m(metal) * c(metal) * ΔT(metal) = - m(water) * c(water) * ΔT(water)
21 grams * c(metal) *(19-122.5) = -100 * 4.184 * (19-17)
-2173.5 *c(metal) = -836.8
c(metal) = 0.385 J/g°C
The metal has a heat capacity of 0.385 J/g°C
This metal is copper.
The term which is used is homogeneous.
when sugar is completely dissolved in the water, the mixture or solution homogeneous, both in same phase and same uniform texture that is liquid.
There two types of mixtures are heterogeneous and homogeneous in different phases.
If sugar is not completely dissolved in water and you see the crystals of sugar in water, then the solution will be heterogeneous.
